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Buttery's Series of Mathematics.
ELEMENTS
OP
PLANE GEOMETRY.
BY
FRANKLIN IBACH, B. S.,
TEACHER OF MATHEMATICS IN THE PEIRCE COLLEGE OF BUSINESS.
PHILADELPHIA; E. H. BUTLER & CO.
0 ■ Z3
Entered, according to Act of Congress, in the year 1882, by
E. H. BUTLER & CO.
In the Office of the Librarian of Congress, at Washington, D. C.
ELECTROTYI'KD BY MACKELLAR, SMITHS 4 JORDAN, PHILADELPHIA.
CONTENTS,
INTRODUCTION.
PAGE
Subject-Mattek 7
BOOK I.
Definitions 9
Mathematical Terms 14
Axioms 14
Postulates 15
Demonstration 15
Symbols and Abbreviations 16
Perpendicular and Oblique Lines .... 17
Parallel Lines 27
Equality of Angles .... . . . .31
Triangles 34
Relation between the Sides of a Triangle . . 36
Medial Lines 38
Angles of a Triangle ....... 39
Equality of Triangles 41
Relation between the Parts of a Triangle . . 48
Bisectors of Angles 51
Polygons . . . 54
Angles of a Polygon 56
Quadrilaterals 58
Parallelograms 60
Exercises in Invention 71
3
CONTENTS.
BOOK II. RATIO AND PROPORTION.
PAGE
Definitions 74
Theorems 77
BOOK III.
THE CIRCLE.
Definitions 85
Chords, Arcs, etc 87
Relative Position of Circles 97
Measurement of Angles 99
Problems in Construction 107
Exercises in Invention 124
BOOK IV. AREA AND RELATION OF POLYGONS.
Definitions 126
Areas 127
Squares on Lines 133
Projection 135
Proportional Lines 140
Similarity of Polygons 143
Relation of Polygons 160
Problems in Construction 164
Exercises in Invention 177
BOOK V. REGULAR POLYGONS AND THE CIRCLE.
Definitions 179
Relation between the Circumference and the
Diameter of a Circle 183
Problems in Construction 188
Exercises in Invention 195
PREFACE
This little volume has been prepared with a view to furnish a suitable text-book on Plane Geometry for Grammar Schools, Preparatory Schools, etc.
A simple method of designating angles has been adopted, and recognized symbols have been freely used in the demon- strations, thus bringing the several steps closely together and enabling the student to master the argument with ease. The reasons on which the steps of an argument depend are not formally given, but are referred to by numbers indicating the sections in which they are found : it is believed that the pupil will impress the principles most firmly on his mind by fre- quency of reference.
No valid objection can be offered against the algebraic form of which some of the demonstrations partake, for most of the axioms laid down are nothing more than properties of the equation.
No apology is deemed necessary for the application of the Infinitesimal method: it has been employed whenever it gave directness, brevity, and simplicity to the demonstration.
At the close of each book, except the second, a collection of theorems and problems has been placed for the purpose of giving the pupil an opportunity to exercise his originality in demonstration and construction. A proper use of these exer- cises will do much toward stimulating thought and awakening a spirit of invention in the pupil.
6 PREFACE.
During the preparation of this treatise, Diesterweg's "Ele- mentare Geometrie" and most of our American treatises have been freely consulted.
And now, this little work is respectfully submitted to the edu- cational public, in the hope that it may at least merit a careful perusal.
F. IBACH.
Philadelphia, Pa., May, 1SS2.
NOTE TO THE TEACHER.
In recitation, when studying a book for the first time, the pupil should be required to draw the diagram accurately and write the demonstration neatly on the blackboard.
Being called upon to recite, he enunciates the proposition and gives the demonstration, pointing to the parts of the diagram as reference is made to them.
In review, the diagram only should be put on the board.
INTRODUCTION.
SUBJECT-MATTER.
The accompanying diagram represents a block of granite, — a physical solid, of regular form.
Such a block has six flat faces, called Surfaces. It has also twelve sharp edges in which these surfaces meet, called Lines.
It has, besides, eight sharp corners in which these lines meet, called Points.
If the block be removed, we can imagine the space which it filled to have the same shape and size as the block. This limited portion of space, which has lefigth, breadth, and thickness, is called a Geometrical Solid. Its boundaries or surfaces separate it from surrounding space, and have length and breadth but no thickness. The boundaries of these surfaces are lines, and have length only. The limits of these lines are points, and have position only. We thus come in three steps from solids to points, which have no magnitude. Having thus acquired notions of solids, surfaces, lines, and points, we can easily conceive of them distinct from one another. It is of such ideal solids, surfaces, lines, and points that Geometry treats; and these in various forms, except points, are called Geometrical Magnitudes or Magnitudes of Space.
ELEMENTS
OF
PLAICE GlEOMETRT.
BOOK^"""
DEFIlSriTIO
1. Geometry is the science which treats of the properties and relations of magnitudes of space.
Space has extension in all directions; but for the purpose of determining the size of portions of space, we consider it as having three dimensions, namely, length, breadth, and thickness.
2. A Point is position without size.
3. A ijine is that which has but one dimension, namely, length.
A line may be conceived as traced by a moving point. Lines are straight or curved,
4. A straight Line is one which has the same direction at all its points.
5. A Ourvea Line is one which changes its direction at all its points. When the sense is ob- vious, the word line, alone, is used for straight line, and the word curve, alone, for curved line.
6. A Surface is that which has only two dimensions, length and breadth.
A surface may be conceived as generated by a moving line. Surfaces are plane or curved,
9
10 ELEMENTS OF PLANE GEOMETRY.
7. A Plane Surface, or a Platte,
is a surface ^vith which a straight line can be made to coincide in any direction.
8. A Curved Surface is a surface no portion of which is a plane.
9. A Solid is that which has three dimensions, lengthy breadth, and thickness. A solid may be conceived as generated by a moving surface.
Points, lines, surfaces, and solids are the concepts of Geometry, and may be said to constitute the subject-matter of the science.
10. A Figure is some definite form of magnitude.
11. Lines, surfaces, and solids are called figures when reference is had to their form.
12. A Plane Figure is one, all of whose points are in the same plane.
13. Plane Geometry treats of plane figures.
14. Equal Figures are such 'as have the same form and size, that is, such as fill exactly the same space.
15. FQuivalent Figures are such as have equal magnitudes.
16. Similar Figures are such as have the same form, although they may have difierent magnitudes.
17. A Plane Angle, or an Angle, is the opening between two lines which meet each other. The point in which the lines meet is called the Vertex, and the lines are called the sides of the angle. A plane angle is a species of surface.
An angle is designated by placing a letter at each end of its sides, and one at its vertex, or by placing a small letter in it near the vertex. The latter is the method employed in this book, whenever it is convenient. In reading, when there is but one angle, we may name the letter at the vertex; but when there are two or more vertices at the same point, we
ELEMENTS OF PLANE GEOMETRY.
11
name the three letters, •with the one at the vertex between the other two: we may, however, in either case, simply name the letter placed in it. Thus, in Fig. 1, we say angle C, or angle a.
Fig. 2.
G
-B
D
Q
E
In Fig. 2, G being tiie common vertex, we must say angle D GF, or angle b. The size of an angle depends upon the exte^it of opening of its sides, and not upon the length of the sides.
18. Adjacent Angles are SUch as have a common vertex and one common side between them. Thus, the angles a and b are ad- jacent angles.
19. A Right Angle is an angle included between two straight lines which meet each other so as to
make the adjacent angles equal.
Thus, if the angles a and b are equal, each is a right angle.
20. Perpendicular Lines are SUch as make right angles with each other.
21. An Actite Angle is one which is less than a right angle; as angle a.
22. An Obtuse Angle is one
which is greater than a right angle; as angle ABC.
Acute and obtuse angles are called oblique angles.
12
ELEMENTS OF PLANE GEOMETRY.
23. OhliQue ijinea are lines which are not perpendicular to each other, and which meet if sufficiently produced.
24. Two angles are Conijplements of each other when their sum is equal to a right angle. Thus, angle a is the com- plement of angle b, and angle b is the complement of angle a.
25. Two angles are Supple- nients of each other when their sum is equal to two right angles. Thus, angle a is the supplement of angle ABC, and angle ABC is the supplement of angle a.
26. Vertical Anyles are such as have a common vertex, and their sides lying in opposite directions. Thus, angles a and b are vertical ; also angles c and d.
-JU- c/d
.A
27. If two lines are cut by a third line, eight angles are formed, which are named as follows:
Angles a, b, c, and d are Exterior
Angles. Angles e, /, g, and h are Interior Angles, The pairs of an- gles a and d, b and c, are Alternate Exterior Angles. The pairs of an- gles e and h, f and g, are Alternate Interior Angles. The pairs of angles a and g, b and h, e and c, / and d, are Cor- responding Angles,
28. Parallel Straight Lines are
such as lie in the same plane and
cannot meet hoAV far soever they '~^
are produced either way. They have the same direction.
ELEMENTS OF PLANE GEOMETRY. 13
29. A Circle is a plane figure bounded by a curve, all the points of which are equally distant from a point within, called the Centre.
The Circumference of a circle is the curve which bounds it.
A Radius of a circle is a line extend- ing from the centre to any point in the circumference.
• The diagram represents a circle whose
centre is 0. The curve ABCD is the circumference, and
the line OA is a radius.
14 ELEMENTS OF PLANE GEOMETRY.
DEFINITIONS OF MATHEMATICAL TEEMS.
30. A netnonstration, or Proof, is a course of reasoning by which the truth of a statement is deduced.
31. An Axiotn is a statement of a truth which is self- evident.
32. A Thcoretn is a statement of a truth which is to be demonstrated.
33. A JProbietn is a statement of something to be done.
34. A Postulate is a problem whose solution is self-evident.
35. Axioms, theorems, and problems are called Propo- sitions.
36. A Corollary is a statement of a truth which is a direct inference fr(jm a proposition.
37. An Hypothesis is a supposition made in a proposition or in a demonstration.
38. A Scholium is a comment on one or more propositions.
39. AXIOMS.
1. Things which are equal to the same thing are equal to each other.
2. If equals are added to equals, the sums are equal.
3. If equals are subtracted from equals, the remainders are equal.
4. If equals are added to unequals, the sums are unequal.
5. If equals are subtracted from unequals, the remainders are unequal.
6. If equals are multiplied by equals, the products are equal.
7. If equals are divided by equals, the quotients are equal.
8. The whole is greater than any of its parts.
9. The whole is equal to the sum of all its parts.
ELEMENTS OF PLANE GEOMETRY. 15
10. Only one straight line can join two points.
11. A straight line is the shortest distance from one point to another.
12. All right angles are equal.
40. POSTULATES.
1. A straight line can be drawn joining any two points.
2. A straight line can be produced to any length.
3. From the greater of two straight lines, a part can be cut equal to the less.
4. In a plane a circumference of a circle can be described, with any point as a centre, and any distance as a radius.
5. Figures can be freely moved in space without change of form or size.
41. DEMONSTRATION.
A Demonstration is a logical process, the premises being definitions and self-evident and previously established truths.
There are two methods of demonstration, called the JOirect MetHod and the indirect Method,
The nirect Method proves a truth by referring to defi- nitions and self-evident and previously deduced truths, and concludes directly with the proof of the truth in question.
The Indirect Method proves a truth by showing that a supposition of its falsity leads to an absurdity; — called also redudio ad absurdum.
Pupils frequently fall into errors of demonstration. Notable among these errors are Begging the Question and Reason- ing in a Circle,
Begging the Qt€estion is a form of argument in which the truth to be proved is assumed as established.
Reasoning in a Circle is a form of argument in which a truth is employed to prove another truth on which the former depends for its proof
16 ELEMENTS OF PLANE GEOMETRY.
42. EXPLANATION OF SYMBOLS AND ABBREVIATIONS.
= denotes equality.
|
+ |
« |
increased by. |
|
— |
ii |
diminished by. |
|
X |
li |
multiplied by. |
|
-f- |
(I |
divided by. |
|
(( |
therefore. |
|
|
II |
it |
parallel. |
|
lis |
ii |
parallels. |
|
z. |
<( |
angle. |
|
As |
t( |
angles. |
|
L |
(C |
right angle. |
|
Ls |
(I |
right angles. |
|
-L |
(C |
perpendicular. |
|
J_s |
il |
perpendiculars. |
|
> |
(( |
is greater than. |
|
< |
(t |
is less than. |
|
A |
li |
triangle. |
|
As |
ii |
triangles. |
|
RA |
(I |
right-angled triangle. |
|
RAs |
ii |
right-angled triangles. |
|
O |
ii |
circle. |
|
Os |
ii |
circles. |
|
o |
ii |
parallelogram. |
|
Os |
ii dene |
parallelograms. |
|
Ax. |
>tes axiom. |
|
|
Cor. |
(( |
corollary. |
|
Cons |
(( |
construction. |
|
Hyp. |
(( |
hypothesis. |
|
Q.E.D. " |
quod erat demonstrandum (which was to |
|
|
be demonstrated). |
||
|
Q.E.F. " |
quod erat faciendum (which was to be done). |
ELEMENTS OF PLANE GEOMETRY. 17
PEEPET^DICULAR Al^D OBLIQUE STRAIGHT LINES.
THEOREM I.
43. At a point in a straight line, only one perpendicular can be erected to that line.
Let O be a point in the line AB.
E D
A a Bs
To prove that only one JL can he erected to AB at C.
Draw the oblique line CD.
Revolve CD about C so as to increase Z. a and decrease Z. ACD.
It is evident that in one position of CD, as EC, the adjacent Z-s are equal.
But then EC is JL to AB. (20)
And there can be only one position of the line in which the adjacent Z_8 are equal;
. • . only one _L can be erected to AB at the point C. Q. E. D.
2*
18
ELEMENTS OF PLANE GEOMETRY,
THEOREM II.
44. The sum of the two adjacent angles formed by two lines which meet equals two right angles.
Let Z.S a and ACD be formed by the line CD meet- ing AB.
0
B
To prove that Z. a + Z_ ^C'i) = 2 Ls. Let CE be _L to AB at C.
L.a^ jL ECD = a L, and /L ACD— Z.ECD = 2iL.
Add ; then /La -^ Z. ACD = 2Lb. (Ax. 2) Q. E. D.
45. Cor. 1. — If one of the two adjacent angles formed by two lines which meet is a right angle, the other is also a right angle.
46. Cor. 2. — The sum of all the angles formed at a common point on the same side of a straight line equals two right angles.
Thus, Z.a4-Z.6+Z.c + Z. cZ + Z.e = 2Ls.
ELEMENTS OF PLANE GEOMETRY. 19
THEOREM III.
47. Conversely. — If the sum of two adjacent angles equals two right angles, their exterior sidts lie in the same straight line.
Jjet Z. a -^ Z. ACD = 2Ls.
E
b/a
0
To prove that A C and B C lie in the same straight line.
Draw EC.
li EG and BC lie in the same straight line,
Z. a + Z. 6 == 2 Ls. (44)
But Z-a-^ I. AGD=2L9', (Hyp.)
Z. a-\-Z-h = I. a-\-l.ACD. (Ax. 1)
From each member subtract Z. a.
Then Z. 6 = Z_ ^ CD, which is impossible.
(Ax. 8) . * . AC and B C lie in the same straight line. Q. E. D.
20 ELEMENTS OF PLANE GEOMETRY.
THEOREM IV.
48. If two straight lin6s intersect, the opposite or vertical angles are equal.
Let AB and CD intersect.
C
To prove that Z- a=^ /- h.
Z. a 4- ^ c = 2 Ls, (44)
and Z. 5 + A c = 2 Ls; (44)
/^a-]rl-c=l.h^A.c. (Ax. 1) From each member subtract Z_ c.
Then /L a = /L h.
Likewise we can prove that L. c = I- d. Q. E. D.
49. Cor. 1. — Ij two straight lines intersect, the sum of the four angles formed equals four right angles.
50. Cor. 2. — The sum of all the angles that can be formed at a common point equals four right angles.
ELEMENTS OF PLANE GEOMETRY,
21
THEOREM V.
51. From a point without a straight line, only one perpendic- ular can he drawn to that line. .
Let P be a point without AB. P
D
C
To prove that only one _L can he drawn from Pto AB.
Draw the oblique line PC. With the point P fixed, revolve PC so as to decrease Z. a and increase Z_ h, while the common vertex moves in the direction CA.
At some position of the line, as PD, the adjacent angles are equal.
Then PD is JL to AB. (20)
There is only one position of the line in which the angles are equal.
only one _J_ can be drawn from P to AB. Q. E. D.
22
ELEMENTS OF PLANE GEOMETRY.
THEOREM VI.
52. From a point without a straight line, a perpendicular is the shortest distance to that line.
Let AB be a straight line, C any point without it, CE a J_, and CF any oblique line.
F\ E
-B
D
To prove that C E < CF.
Produce CE to D, making ED = CE, and draw FD.
On AB as an axis, revolve the plane of CEF till it falls in the plane of DEF.
Since Z. s a and h are Ls, the line CE takes the direction ED, the point C falling on D.
or
CE=ED; CE-ir ED = 2 CE.
CF=FD; CFA- FD = 2 CF. CE+ ED< CF-]- FD, 2 CE<:2 CF.
Divide by 2 ; then
CE < CF.
(Cons.) (Ax. 10)
(Ax. 11) Q.E.D.
ELEMENTS OF PLANE GEOMETRY.
23
THEOREM VII.
63. Any point in the perpendicular erected at the middle point of a straight line is equally distant from the extremities of that line.
Let P be any point in CD which is JL to AB at its middle point D, and let ^P and BP be drawn.
To prove that AP = BP.
On CD as an axis, revolve APD till it falls in the plane of BPD.
Since Z_s a and h are Ls, and AD AP = BP.
BD, A falls on B. (Ax. 10) Q. E. D.
54. Cor. 1. — If a point is equally distant from the extremi- ties of a straight line, it lies in the perpendicular erected at the middle point of that line.
55. Cor. 2. — If each of two points in a straight line is equally distant from the extremities of another straight line, the former is perpendicular to the latter at its middle point.
24
ELEMENTS OF PLANE GEOMETRY.
THEOREM VIII.
56. Any point without the perpendicular erected at the middle point of a straight line is unequally distant from the extremities of that line.
Let P be any point without the line CD which is J_ to AB at its middle point. Draw AP cutting CD at 0, and draw^O and ^P.
To prove that AP > PP.
BO + OP>BP, (Ax. 11)
and AO = BO. (53)
Substitute A 0 for its equal B 0. Then A0-\- OP > BP.
But AO-{^OP=AP;
AP>BP. Q.E.D.
THEOREM IX.
57. Two oblique lines drawn from a point in a perpendicu- lar are equal if they cut off equal distances from the foot of the perpendicular.
C
Let CD be J_ to AB, and CE and (7P oblique lines cut- ting off ED = DF.
To prove that CE = CF. ^
On CD as an axis, revolve CDE till it falls in the plane of CDF.
Since /.g a and b are Ls, and ED = DF, E falls on F. CE=CF. (Ax. 10) Q.E.D.
ELEMENTS OF PLANE GEOMETRY. 25
THEOREM X.
58. The sum of two lines drawn from a point to the extremi- ties of a straight line is less than the sum of two other lines similarly drawn and enveloping them.
Let AP and BP be two lines drawn from P to the extremi- ties oi AB, and let AC and BChe two lines drawn similarly and enveloping AP and BP.
To prove that AP ^ BP < AC + BC.
Produce AP to D, a point in BC.
AP ■\- PD < AC -\- CD, and BP<PD-\- DB. (Ax. 11)
Add the inequalities. Then AP-\-BP-^PD <AC^ CD ^ DB-^PD.
Substitute BC for its equal CD + DB, and subtract PD from each member.
Then AP-\-BP<AC^Ba Q. E. D.
26
ELEMENTS OF PLANE GEOMETRY.
THEOREM XI.
59. Of two oblique lines drawn from the same point, that is the greater which terminates at the greater distance from the foot of the perpendicular.
Let CO be J_ to AB, and CE and CD oblique lines drawn so that EO > DO.
E\I)\
To prove that CE > CD.
Produce CO to F, making OF = CO, and draw EF andDF.
Then, as in (52), CD = DF, and CE = EF.
But CE-\-EF> CD + DF,
or 2 C^ > 2 CD. (58)
Divide each member by 2.
Then CE > CD. Q. E. D.
60. Cor. 1. — Two equal oblique lines terminate at equal distances from the foot of the perpendicular.
61. Cor. 2. — Only two equal straight lines can be drawn from a point to a line; and of two unequal oblique lines, the greater terminates at the greater distance from the foot of the perpendicular.
ELEMENTS OF PLANE GEOMETRY. 27
PAEALLEL STKAIGHT LIInTES.
THEOREM XII.
62. If two parallel lines are cut by a third line —
I. The corresponding angles are equal;
II. Tlie alternate interior angles are equal;
III. The sum of the interior angles on the same side of the secant equals two right angles.
Let the lis AB and CD be cut by the line EF.
E
A 0/a T>
A yt" B
F
I. To prove that Z_ a = Z. &.
Since OB and PD are II, they have the same direction and open equally from the line EF;
Z.a=^b.
Likewise we can prove that /L c = /- d.
II. To prove that /- c = Z^ b.
^c = Z.a. (48)
But A. a = Z.b; (Case I.)
/.c = Z-b. (Ax. 1) Likewise we can prove that /^ e = Z. f.
28 ELEMENTS OF PLANE GEOMETRY.
III. To prove that Z. b -\- Z. e = 2Ls.
A a + Z. e = 2 Ls. (44)
Substitute Z- b for its equal Z_ a.
Then Z.6 +Z- e = 2L.. Likewise we can prove that Z. c -j- /^ f = 2Ls. Q. E. D.
63. Cor. — If a straight line lies in the same plane with two parallels, and is perpendicular to one of them, it is perpendicu- lar to the other also.
THEOREM XIII.
64. If two straight lines are cut by a third line, these two lines are parallel —
I. If the corresponding angles are equal;
II: If the alternate interior angles are equal;
III. If the sum of the two interior angles on the same side of the secant equals two right angles.
Let the straight lines AB and CD he cut by EF.
E
F
I. To prove that AB and CD are II if I- a=^ A^b.
If /^a = Z-b, OB and FD open equally from EF, and hence have the same direction, or are 11. (28)
II. To prove that AB and CD are W if Z^ d = Z. b.
If Z.a = Z.b, AB and CD are II. (Case I.)
But Z.d = Z. a; (48)
if Z.d = Z.b, AB and CD are II.
ELEMENTS OF PLANE GEOMETRY.
29
III. To prove that AB and CD are W if Z, c -^ Z. b = 2 Ls.
l.a-{- Z-c = 2U; Z^c -\- Z. b = 2Ls; Z.a-{-Z.c=Z-G-{-Z-b.
Subtract Z_ c from each member. Then Z. a = Z. b.
But then the lines are II ; . if /. c -f Z. 6 = 2 Ls, ^^ and CD are II
(44) (Hyp.)
(Ax. 1)
(Case I.) Q.E.D.
65. Cor. — If two lines in the same plane are perpendicular to the same line, they are parallel.
THEOREM XIV.
66. If two straight lines are parallel to a third line, they are parallel to each other.
Let AB and CD be
to EF. G
H
To prove that AB is \\ to CD. Let GH be JL to EF; Then OH is JL to AB. GH is also JL to CD; . AB is !I to CD.
(63)
(63)
(65) Q.E.D.
30 ELEMENTS OF PLANE GEOMETRY.
THEOREM XV.
67. Two parallel lines are everywhere equally distant from each other.
Let^jB and CD be 11, and GO and jffP-Ls to CD drawn from any two points in AB.
To prove that GO = HP.
Let the _-L EF be drawn at the middle point between G and H.
GO, HP, and EF are _L to both lis. (63)
On EF as an axis, revolve the part of the plane on the right of EF till it falls in the part on the left of EF.
Z.S a and h are Ls, and EH = EG;
/f falls on G^.
Z- s c and d are Ls ; .
.-. HP takes the direction GO, and P falls in G^O, or GO produced.
Z.8 m and n are Ls; P falls in the line FC
Now P falling in both GO and FC, must fall at their intersection 0;
GO=HP. Q.E.D.
ELEMENTS OF PLANE GEOMETRY.
31
EQUALITY OF AI^TGLES.
THEOREM XVI.
68. Two angles are equal if their sides are respectively parallel and lying in the same or in opposite directions from their vertices.
I. Let AB and BC, the sides of Z. a, be respectively II to DE and EF, the sides of Z. 6.
L
To prove that Z. a = Z. 6.
If necessary, produce two sides not II till they intersect, as at G.
|
Then |
/- a = /- c, |
|
|
and |
^b= Z.c; |
(62) |
|
.-. |
L.a= /_h. |
(Ax. 1) |
II. Let LM and ifiV, the sides of Z. d, be respectively 11 to PQ and OP^ the sides of Z_ e.
To prove that /- d = /- e.
If necessary, produce two sides not II till they intersect, as ati?.
|
Then |
^d=L,f, |
|
|
and |
Z.e= A.f; |
(62) |
|
l.dr= l_e. |
(Ax. 1) Q. E. D. |
32 ELEMENTS OF PLANE GEOMETRY.
THEOREM XVII.
69. Two angles are supplements of each other if they have two sides parallel and lying in the same direction, and the other sides parallel and lying in opposite directions from their vertices.
Let Z.S a and e have AC and EF II and lying in the same direction, and AB and ED II and lying in opposite directions from their vertices.
To prove that Z. s a and e are supplements of each other.
If necessary, produce two sides not II till they intersect, as at G.
Then Z. a = Z. 5, and Z. ci = Z. 6. (62)
But Z. s 6 and d are supplements of each other ;
Z.S a and e are supplements of each other.
Q. E. D.
ELEMENTS OF PLANE GEOMETRY.
33
THEOREM XVin.
70. T\vo angles having their sides respectively perpendicular to each other are either equal or supplements of each other.
Let the Z.S a and b have the sides BA and BC respec- tively J- to EF^ud ED.
M 0 A
|
\ |
/ |
||
|
\ |
|||
|
\ |
X |
||
|
N \ |
/ |
||
|
\ \ |
/ |
||
|
\ \c |
e, / |
||
|
\^ |
Ya |
||
|
5 E |
|||
|
\ |
|||
|
'\^ |
h\ |
||
|
^\ |
\ |
||
|
d |
G |
\ |
|
|
I |
) |
> |
To prove that Z. s a and b are either equal or supplements of each other.
At B, let BO and BM be drawn respectively II to ED and EF, and let G^iV' be drawn II to ^i^ or BM.
Then I. c = L.b, (68)
and Z. c is the supplement of Z. d. (69)
Now Z-c-{-Z^e = ?i\^, and Z_e-|- Z_a = aL;
Z-c -f-Z-e=Z_e+ /.a.
Subtract Z. e from each member.
Then /L c = /^ a.
But Z. c= I- b;
Z.a=Z-b.
Also, since Z. c is the supplement of Z_ d^
Z. a is the supplement of Z. c?. Q. E. D.
34
ELEMENTS OF PLANE OEOMETBY
TEIANGLES.
DEFINITIONS.
71. A Triangle is a plane figure bounded by three straight lines; as ABC.
The Sides of a triangle are the bounding lines.
The Angles of a triangle are the an| formed by the sides meeting one another. A triangle has six parts, — three sides and three angles.
The nase of a triangle is the side upon which it is sup- posed to stand.
The Vertical Angle of a triangle is the angle opposite the base. An Exterior Angle of a triangle is an angle formed by a side and an adjacent side produced ; as Z_ a.
The Vertex of a triangle is the angular point at the ver- tical angle.
The Aititudti of a triangle is the perpendicular distance
from the vertex to the base, or to the base produced. Thus, CD is the altitude of both the triangles ABC and EBC
A Medial JLine of a triangle is a line drawn from a vertex to the middle of the opposite side.
72. Triangles are classified sides and angles.
to their
E B
ELEMENTS OF PLANE GEOMETRY.
35
A Scaietie Triangle is one which has no two sides equal. An Isosceles Triangle is one which has two sides equal.
An Equilateral Triangle is one which has all its sides equal.
An Acute- Angled Triangle is one which, has three acute angles.
A uigHt- Angled Triangle is one which has one right angle.
An Obtuse- Angled Triangle is one which has one obtuse angle.
An Equiangular Triangle is one which has all its angles equal.
36
ELEMENTS OF PLANE GEOMETRY.
RELATION BETWEEN THE SIDES OF A TRIANGLE.
THEOREM XIX.
73. Any side of a triangle is greater than the difference between the other two sides.
Let ABC be any A.
To prove that AC> AB — BC.
AB<AC-\-Ba Subtract BC from each member. Then AB — BC<AC,
or
AOAB—BC.
(Ax. 11)
Q.E.D.
ELEMENTS OF PLANE GEOMETRY.
37
THEOREM XX.
74. The sum of the three lines drawn from a point tuithin a triangle to the vertices is greater than half the sum of the sides of the triangle.
Let m, n, o, be the three lines drawn from any point P in the ^ ABC, and a, b, c, the sides of the A.
To prove that m -^ n -\- o > ^
m + 0 > a,
n -\- 0 ^ c,
and m -\- n'> b.
Add these inequalities.
Then 2m -^ 2n -\- 2o > a -{- b -\- c,
a -\- b -\- c
(Ax. 11)
or
m -|- n + 0 >
Q.E.D.
38
ELEMENTS OF PLANE GEOMETRY.
MEDIAL LINES.
THEOREM XXI.
75. The sum of the three medial lines of a triangle is greater than half the sum of the sides of the triangle. q
|
In the Z\ ABC, let a, b, c, be |
/ |
|
he sides, and m, n, o, the medial ines. A A |
/ 0^ |
|
. To prove that m -\- n -{- o ^ ^ |
|
|
m> a — 2 |
|
|
m>6-2 |
|
|
^ h |
|
|
> |
|
|
<'>*-L |
Add these inequalities.
Then 2m + 2n + 2o > a + 6 + c,
a + 6 -f- c
(73)
or
m + n -f- 0 >
Q.E.D.
ELEMENTS OF PLANE GEOMETRY. 39
AI^GLES OF A TRIANGLE.
THEOREM XXII.
76. The sum of the three angles of a triangle equals two right angles.
0 D
Let ^^C be any A.
B F
To prove that Z^ a -{- Z- h -\- Z^ c ^= 1 \_s.
Produce AB, and let DBhQ I! to AC. Then A m + Z. w + A a = 2 Ls. ^ (46)
But jL m=^ Z-h,
and Z- n = Z^ c. (62)
For Z. 8 m and n, substitute their equals Z. s 6 and c. Then Z. a + Z. 6 -I- ^L c -= 2 Ls. Q. E. D.
77. Cor. 1. — The sum of two angles of a triangle being given, the third can he found by subtracting their sum from two right angles.
78. Cor. 2. — If two angles of a triangle are respectively equal to two angles of another y the third angles are also equal.
79. Cor. 3. — In any triangle, there can be but one right angle, or but one obtuse angle.
80. Cor. 4. — In any right-angled triangle, the sum of the acute angles equals a right angle; that is, they are complements of each other.
81. Cor. 5. — In an equiangular triangle, each angle equals one-third of two right angles.
40 ELEMENTS OF PLANE GEOMETRY.
THEOREM XXIII.
82. An exterior angle of a triangle equals the sum of the two interior non-adjacent angles.
Let ABC be any A, and a an exterior Z..
C
A B
To prove that Z^a=Z^c-\-/-d.
Z. a + ^ 6 = 2 Ls, . (44)
and Z. 6 + ^ c + Z. cZ =: 2 Ls; (7.6)
l.a-^/Lh=Z-h^A.c-^Z.d. Subtract Z_ h from each member ; then Z. a = Z. c + Z. d Q. E. D.
ELEMENTS OF PLANE GEOMETRY.
41
EQUALITY OF TEIANGLES.
THEOREM XXIV.
83. Two triangles are equal in all their parts if two sides and the included angle of the one are respectively equal to two sides and the included angle of the other.
In the As ABC and DEF, let AB =DE, AC = DF, and /- a = /- b .
A B r>
To prove that A ABC = /^ DEF.
Place the A ABC upon DEF, applying Z. a to its equal Z. h, AB on its equal DE, and AC oti its equal DF.
The points C and B fall on the points F and E;
BC=EF. (Ax. 10)
and ^ A ^^C= A DEF. (14)
Q.E.D.
42
ELEMENTS OF PLANE GEOMETRY,
THEOREM XXV.
84. Two triangles are equal in all their parts if two angles and the included side of the one are respectively equal to two angles and the included side of the other.
In the As ABC and DEF, let Z. a = Z. c?, Z.6 = Z. e, and AB = DE.
To prove that A ABC= Z^ DEF.
Place the A ABC upon the A DEF, applying AB to its equal DE, the point A on D, and the point B on E.
Since Z. a = /- d, AC takes the direction DF, and O falls somewhere in DF or DF produced.
Since Z. b = Z. e, BC takes the direction EF, and C falls in EE' or EF produced;
. • . the point C, falling in both DF and EF, falls at their intersection F;
A ABC = A DEF. (14) Q. E. D.
85. Cor. — If a triangle has a side, its opposite angle, and one adjacent angle, respectively equal to the corresponding parts of another triangle, the triangles are equal.
ELEMENTS OF PLANE GEOMETRY. 43
THEOREM XXVI.
86. Two triangles are equal in all their parts if the three sides of the one are respectively equal to the three sides of the other.
In the As ABC and DEF, let AC = DF, BC = EF, and AB = DE.
To prove that A ABC = ^ DEF.
Place ABC in the position DEG, AB in its equal DE, and the Z.s a and c adjacent to the Z_ s 6 and d.
Draw FO cutting DE at P.
DG = DF, and EG = EF; (Hyp.)
the points D and E are equally distant from F and G, and DE is JL to EG at its middle point. (55)
On DE as an axis, revolve DEG till it falls in the plane of DEF.
Then the point G falls on jP, since PG = PF;
A ABC = A DEF. (14) (Q. E. D.)
87. Cor. — In equal triangles, the equal angles lie opposite the equal sides.
88. Scholium. — The statement, two triangles are equal, means that the six parts of the one are respectively equal to the six parts of the other.
44
ELEMENTS OF PLANE GEOMETRY.
THEOREM XXVII.
89. If two triangles have two sides of the one respectively equal to two sides of the other, and the included angles unequal, the third sides are unequal, ayid the greater third side is in the triangle having the greater included angle.
In the As ABC and DEF, let ^C = DF, CB = FE,
and ACB > Z. c.
To prove that AB > DE.
Place the A DEF so that EF falls in its equal BC. Let CH bisect Z. ECA, and draw EH.
CE=CA, (Hyp.)
CH=- CH, and /- a = A. b; (Cons.)
ACHE= A CHA, (83)
and EH=AH.
Now HB + EH> EB.
Substitute AH and DE for their equals EH and EB. Then AH-^ HB> DE,
or
AB > DE.
Q.E.D.
ELEMENTS OF PLANE GEOMETRY.
45
THEOREM XXVin.
90. Conversely. — If two sides of a triangle are respectively equal to two sides of another, and the third sides unequal, the angle opposite the third side is greater in the triangle having the greater third side.
In the As ABC and DEF, let AC=DF, BC = EF, and AB > DE.
To prove that /L a'> Z. 6.
If Z. a = 21 6, A ABC = A DEF, (83)
and AB = DE.
If /.a< Z- b,AB< DE. (89)
But both these conclusions, being contrary to the hypoth- esis, are absurd.
. • . /-a cannot equal Z- b, and cannot be less than Z. b, Z.a> ^b. Q.E.D.
46
ELEMENTS OF PLANE GEOMETRY.
THEOREM XXIX.
91. Two right-angled triangles are equal in all their parts if the hypothenuse and one side of the one are respec^vely equal to the hypothenuse and one side of the other.
In the RAs ABC and DEF, let the hypothenuse BC = EF. and AC =DF.
D
E
To prove that R/\ ABC = BA DBF.
Place ABC on DBF so that ^C falls in its equal DF.
Z-B a and b are Ls; (Hyp.)
AB takes the direction DE, and B falls on E; (60)
R A ABC = R A DBF. (14) Q. E. D.
ELEMENTS OF PLANE GEOMETRY.
47
THEORE
92. Two right-angled triangles are equal in all their paris if the hypothenuse and one acute angle of the one are respectively equal to the hypothenuse and one acute angle of the other.
In the RAs ABC and DEF, let BC = EF, and ^ a = Z. b.
To prove that i?A ABC=EZ^ DEF.
and
|
BC=EF, |
(Hyp.) |
|
/_a= l.h, |
(Hyp.) |
|
Z.c= Ad; |
(78) |
|
RA ABC=B,/\ DEF. |
(84) Q.E.D. |
93. Cor. — Two right-angled triangles are equal if a side and an acute angle of the 07ie are respectively equal to a side and an acute angle of the other.
48 ELEMENTS OF PLANE GEOMETBT.
RELATION BETWEEN THE PAETS OF A TRIANGLE.
THEOREM XXXI.
J-
94. In an isosceles triangle, the angles opposite the equal sides are equal.
In the isosceles A ABC, lei AC and BC he the equal sides.
C
A D B
To prove that L. a =^ L. h.
Let CD be drawn to the middle of AB.
AC=BC, (Hyp.)
AD = BD, (Cons.) and CD = CD;
A ADC= A BDC, (86)
and /. a= Z. b. Q. E. D.
95. Cor. 1. — The straight line joining the vertex and the middle of the base of an isosceles triangle bisects the vertical angle and is perpendicular to the base.
96. Cor. 2. — The straight line which bisects the vertical angle of an isosceles triangle bisects the base at right angles.
97. Cor. 3. — Any equilateral triangle is equiangular.
ELEMENTS OF PLANE GEOMETRY.
49
THEOREM XXXII.
98. Conversely. — IJ two angles of a triangle are equal, the sides opposite them are equal, and the triangle is isosceles.
In the A ABC, let Z. a
To prove that AC = BC.
Let CD be J_ to AB.
/.a:= Z.b, (Hyp.)
and CD = CD;
B.AADC=B.ABDC, (93)
and AC=BC. Q. E. D.
THEOREM XXXIII.
99. Of two angles of a triangle, the greater is opposite the greater side. O.
In the A ABC, let CB > AB.
To prove that Z. BAO L. e. Cut off BD = AB, and draw AD.
Z. a= I- d. (94)
But Z. d> Z. c; > (82)
Z. a> ZL c. And much more is Z. BAO Z. c. Q. E. D.
50
ELEMENTS OF PLANE GEOMETRY.
THEOREM XXXIV.
100. Conversely. — Of two sides of a triangle, the greater is opposite the greater angle.
In the A ABC, let /L BAO Z. c.
To prove that BO AB.
Let AD be drawn so as to make Z-a^= JL c.
Then AD == CD.
Now AD -^ BD> AB.
Substitute CD for its equal AD.
Then CD -\^ BD> AB,
or BO AB.
(98) (Ax. 11)
Q. E. D.
ELEMENTS OF PLANE GEOMETRY.
51
BISECTOES OF ANGLES.
THEOREM XXXV.
101. Any point in the bisector of a?i angle is equally distant from the sides of the angle.
Let BF be the bisector of the Z. ABC, P any point in it, and PD and PE J_s to AB and BC.
To prove that PD = PE.
|
Z_3 a and h are Ls, |
||
|
/-c= jLd, |
(Hyp.) |
|
|
and |
BP = BP; |
|
|
R A BDP = EA BEP, |
(92) |
|
|
and |
PD = PE. |
Q. E. D. |
102. Cor. — Any point in an angle equally distant from the sides lies in the bisector of the angle.
62 ELEMENTS OF PLANE GEOMETRY.
THEOREM XXXVI.
103. The bisectors of the angles of a triangle meet in a common point.
Let AEy BFj and CD be the bisectors of the Z.s of the A ABC.
A B
To prove that AE, BF, and CD meet in a common point.
Let AE and BF meet in a point, as 0. Then 0 is equally distant from AB and AC; also from ABandBC; (101)
.*. 0 is equally distant from J. C and BC, and lies in CD;
(102)
.*. the bisectors AE, BF, and CD meet in a common point.
Q. E. B.
104. Cor. — The point in which the bisectors of the angles of a triangle meet is equally distant from the three sides of the triangle.
ELEMENTS OF PLANE OEOMETBY. 63
THEOREM XXXVII.
105. The perpendiculars erected at the middle points of the sides of a triangle meet in a common point.
In the A ABC, let DH, FG, and EM be respectively _L to AC, AB,aiid BC, at their middle points.
To prove that DH, FG, and EM meet at a common point.
The J_s DH and FG meet in some point, as 0, other- wise they would be II, and AC and AB, the JLs to these lis, would lie in the same straight line, which is impossible.
Now 0 is equally distant from A and C; also from A and B; (53)
.-. 0 is equally distant from C and B, and must lie in EM (54). That is, the J_ EM passes through 0;
. • . DH, FG, and EM meet in a common point. Q. E. D.
106. CoE. — The common point of the perpendiculars erected at the middle points of the sides of a triangle is equally distant from the vertices of the triangle.
5*
54 ELEMENTS OF PLANE GEOMETRY.
POLYGONS.
DEFINITIONS.
107. A roiygon is a plane figure bounded by straight lines. The bounding lines are the sides of the polygon.
The Perimeter of a polygon is the sum of the bounding lines. The angles which the adjacent sides make with each other are the angles of the polygon.
A niagonai of a polygon is a line join- ing two non-adjacent angles.
Note. — Let the pupil illustrate.
108. An Equilateral Polygon, is one all of whose sides are equal.
109. An EquianguUir Polygon is one all of whose angles are equal.
Two polygons are mutually equilateral when their sides are respectively equal.
Two polygons are mutually equiangular when their angles are respectively equal.
Homologous sides or angles are those which are similarly
placed.
110. A Convex Polygon is one no side of which when produced can enter the surface bounded by the perimeter.
Each angle of such a polygon is called a salient angle.
ELEMENTS OF PLANE GEOMETRY.
55
111. A Concave Polygon is one of which two or more sides, when produced, will enter the space enclosed by the perimeter.
1 Convex 1 1 Polygon. 1
The angle ^ OC is called a re-entrant angle.
112. By drawing diagonals from the vertex of any angle of a polygon, it may be divided into as many triangles as it has sides less two.
113. A polygon of three sides is a Triangle; of four, a Quadrilateral; of five, a Pentagon; of six, a Hexagon; of seven, a Heptagon ; of eight, an Octagon ; of nine, a Nona- gon ; of ten, a Decagon ; of twelve, a Dodecagon.
66 ELEMENTS OF PLANE GEOMETRY.
ANGLES OF A POLYGOI^.
THEOREM XXXVIII.
114. The sum of all the angles of any polygon equals two right angles taken as many times less two as the polygon has
Let ABCDEF be a polygon of w sides.
E
^D
To prove that Z. AFE + Z. FED -}- Z. EDO, etc. =
2 Ls (n— 2) From any vertex, as F, draw the diagonals FB, FC, and FD.
Then we have (ti— 2) As. (112)
The sum of the Z.s of the As = the sum of the Z.sof the polygon.
But the sum of the Z.s of a A = 2 Ls; (76)
the sum of the Z_s of the polygon = 2 Ls (n — 2).
Q.E.D. 115. Cor. — The sum of the angles of a quadrilateral is 4 Ls; of a pentagon, 6 Ls ; of a hexagon, 8 Ls, etc.
ELEMENTS OF PLANE GEOMETRY. 57
THEOREM XXXIX.
116. If the sides of a convex polygon are produced so as to form one exterior angle at each vertex, the sum of the exterior angles equals four right angles.
Let ABODE be a polygon oin sides, and let the sides be produced so as to form the exterior Z_ s a, 6, c, cZ, e.
a B
To prove that /.a+Z.5 + Z.c+^c^+/.e = 4Ls.
At each vertex there are two Z.s whose sum = 2 Ls; (44)
and since there are as many vertices as there are sides, we have n X 2 Ls.
But the sum of the interior Z_s ;= 2 Ls (w — 2); (114)
./:^a^l.h-{-/Lc-\-/-d-^A.e=n X 2 Ls —
2 Ls (n — 2)
= 2nLs — 2nL8 + 4Ls.
= 4Ls.
Q.E.D.
58 ELEMENTS OF PLANE GEOMETRY.
QUADEILATERALS.
IDEFINITIOlSrS.
117. A Quadrilateral is a polygon of four sides.
118. There are three classes of quadrilaterals, namely, Trapeziums, Trapeisoids, and Parallelograms.
119. A Trapezium is a quadrilateral which has no two of its sides parallel.
120. A Trapemoia is a quadrilateral which has two of its sides parallel.
The parallel sides are called the bases.
121. A Parallelogram is a quadrilateral which has its opposite sides parallel.
The side upon which a parallelogram is supposed to stand and the opposite side are called the bases.
122. A Rectangle is a parallelogram whose angles are right angles.
123. A SQuare is an equilateral rectangle.
124. A Ritomhoia is a parallelogram whose angles are oblique.
ELEMENTS OF PLANE GEOMETRY.
59
125. A Mhombuft is an equilateral rhomboid.
|
z |
/ Trapeziom |
. |
/ Trapezoid. \ |
|
\ ParaUelogram. \ |
Rectangle. |
||
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\ \ |
|||
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Square. |
\ Rhomboid. \ |
||
|
I |
126. A Diagonal of a parallelogram is a line joining any two opposite vertices.
Note. — Let the pupil illustrate.
127. The Aitituae ;of a parallelogram or trapezoid is the perpendicular distance between its bases.
60 ELEMENTS OF PLANE GEOMETRY.
PAEALLELOGRAMS.
THEOREM XL.
128. In any parallelogram, the opposite sides and angles are equal.
Let ABCD be any CJ.
C D
|
r^"- 71 \ |
h |
|
c |
|
|
/ ^^^-^ |
|
|
/ ^-^ / |
|
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/ "^^ J |
|
|
la >«^-.^/ |
|
|
A |
B |
To prove that AB = CD, AC = BD, Z. a = Z. ^>, and Z_ ACD= Z^ABD.
Draw the diagonal BC.
Z_ 7/1 = Z_ n, (62)
^ c =1. d, (62)
and BC=BC;
A ABC= ^ CDB, (84)
and AB=CD,AC= BD, and Z^ a = Z. b. From the first two equations, '
Z^ c -i- Z^ n = Z^ d -\- ZL m, or Z, ACD^ Z. ABD. Q. E. D.
129. CoK. 1. — A diagonal of a parallelogram divides it into two equal triangles.
130. Cor. 2. — Parallels intercepted between parallels are equal.
ELEMENTS OF PLANE GEOMETRY. 61
THEOREM XLI.
131. Two parallelograms are equal if two sides and the included angle of the one are respectively equal to two sides and the included angle of the other.
Let ABCD and EFGH be two zZ7s, having AD = EH, AB = EF, and Z. a = Z. 6.
C H
B E
To prove that OJ ABCD = CJ EFOH.
Draw the diagonals BD and FH.
AD = EH, (Hyp.)
AB = EF, (Hyp.)
and Z. a=: Z. b; (Hyp.)
A ABD = A EFH (83)
But these As are the halves of the zZ7s.
CD ABCD = cj EFGH Q. E. D.
62 ELEMENTS OF PLANE GEOMETRY.
THEOREM XLII.
132. If the opposite sides of a quadrilateral are equal, the figure is a parallelogram.
In the quadrilateral ABCD, let AB = CD, and A C= BD. C D
a- ^i
A B
To prove that ABCD is a CJ. Draw the diagonal B C.
AB = CD, (Hyp.)
AC=BD, (Hyp.)
and BC=BC;
AABC=^GDB, (86)
and Z. a ^ Z. b;
^J5 and Ci) are 11. (64)
Also, Z. o = Z- d;
AC and BD are W, (64)
and ABCD'm&CJ. (121) Q.E.D.
ELEMENTS OF PLANE GEOMETRY. 63
THEOREM XLIII.
133. If two sides of a quadrilateral are equal and parallel, the figure is a parallelogram.
Let AB CD be a quadrilateral, having AB and CD equal
and II.
|
r\ ■ |
||
|
/ \^j |
||
|
A * B |
||
|
To prove that AB CD is a CD. |
||
|
Draw the diagonal BC. |
||
|
AB = CD, |
(Hyp.) |
|
|
BC=BC, |
||
|
and JL a=^ /-h; |
(62) |
|
|
AABC=^CDB, |
(83) |
|
|
and /^ e= Z. d; |
||
|
^Cand^Darell, |
(64) |
|
|
and ^^ODlsao. |
(121) |
Q.E.D. |
64
ELEMENTS OF PLANE GEOMETRY.
THEOREM XLIV,
134. The diagonals of a parallelogram bisect each other. Let ABCD be a OJ, and AD, B C, its diagonals. C D
A B
To prove that A 0 = DO, and BO = CO.
/^ a =^ /- c,
l^h= Z.d,
and AB=CD;
AABO = ^CDO; whence AO = DO, and BO = CO.
(62)
(62)
(128)
(84)
Q. E. D.
135. Cor. — The diagonals of a rhombus bisect each other at right angles.
ELEMENTS OF PLANE GEOMETRY.
65
THEOREM XLV.
136. The diagonals of a rectangle are equal.
Let AB CD be a rectangle, and J.Z>, B Q its diagonals.
and
and
(128)
(122)
(83)
Q. E. D.
13Y. Cor. — The diagonals of a square are eqxial and bisect each other ai right angles.
oQ ELEMENTS OF PLANE GEOMETRY.
THEOREM XLVI.
138. If a parallel to the base of a triangle bisects one of the sides, it bisects the other also; and the part of it intercepted between the sides equals half the base.
In the A J.^0, let DE be II to the base ^5 and bisect ACoXD.
C
A F
I. To prove that DE bisects BC.
Ijet DF he W to B a
Z. a= Z- c, (62)
Z.b= Z.d, (62)
and AD = DC; (Hyp.)
AADF= A DCE, (84)
and DF = CE.
FBDE is a CJ; (121)
DF=EB. But DF= CE;
EB=CE.
II. To prove that DE = i AB.
In the equal As J.i>Fand DCE, AF=DE, and in the nj FBDE,
DE=FB; (128)
AF= FB, or F is the middle point of AB. DE:=hAB. Q.E.D.
139. Cor. — The straight line which joins the middle points of two sides of a triangle is parallel to the third side and is equal to half that side.
ELEMENTS OF PLANE GEOMETRY. 67
THEOREM XLVII.
140. The parallel to the bases of a trapezoid, bisecting one of the non-parallel sides, bisects the other also.
Let ABCD be a trapezoid, AB and DC its bases, E the middle point oi AD, and let EF be II to AB and DC.
"---.0
A B
To prove that EF bisects B C.
Draw the diagonal BD.
In the A ABD, BD is bisected at 0. (138)
Then in the A DBC,BC is bisected at F; (138)
^F bisects^ a Q.E.D.
141. Cor. T/ie straight line joining the middle points of the non-parallel sides of a trapezoid, is parallel to the base and equals half their sum.
68
ELEMENTS OF PLANE GEOMETRY.
THEOREM XLVIII.
142. The straight line drawn from the vertex of a right angle of a right-angled triangle to the middle of the hypothenuse is equal to half the hypothenuse.
Let the RA ABC be right-angled at B, and let BD be drawn to the middle of A C.
A E
To prove that BD = ^ AC.
Let ED be II to BC.
Z_ 6 is a L.
ED bisects AB;
or, AE = BE,
ED = ED,
and Z- a = /- b ;
aaed = abed,
and BD = AD=^AC.
(63)
(138)
(45)
(83)
Q. E. D.
ELEMENTS OF PLANE GEOMETRY.
69
THEOREM XLIX.
143." The perpendiculars drawn from the vertices of a triangle to the opposite sides meet in a commo7i point.
In the A ABC, let AD, BE, and CF be the vertices to the opposite sides.
from the
To prove that AD, BE, and CF meet in a common point.
Through the vertices, let MR, GM, and GH be drawn respectively 11 to AB, BC, and AC.
ABHC and ABCMave CDs; (121)
AB=CE:==MC, (128)
and C is the middle point of MH.
Kow CF is _L to AB and MH; (63)
CF is _L to 3fH at its middle point.
Likewise we can prove that AD and BE are J_s to GM and GH at their middle points ;
.*. the three -Ls meet in a common point. (105) Q.E. D.
70 ELEMENTS OF PLANE GEOMETRY.
THEOREM L.
144. The medial lines of a triangle meet in a common point. In the A ABC, let AD, BE, and CF be the medial lines. C
To prove that AD, BE, and CF meet in a common point.
Let AD and CF meet at P, and let M and N be the middle points of CP and AP.
Draw MN, NF, FD, MD.
MN is II to ^ 0 and equals h AC, (139)
FD is I! to ^ C and equals I A C; (139)
MN and FD are li (66) and equal,
and NFDM is ncj; (133)
PF= PM (134) =MC=^ CF.
Or AD intersects CF at P, a point whose distance from F equals J OP.
Likewise we can prove that BE intersects CF at a point whose distance from F equals i CF.
AD, BE, and CF meet in a common point. Q. E. D.
ELEMENTS OF PLANE GEOMETRY. 71
EXEECISES 11^ IlSrYENTIOK
THEOREMS.
1. The two straight lines which bisect the two pairs of vertical angles formed by two lines are perpendicular to each other.
2. Two equal straight lines drawn from a point to a straight line make equal angles with that line.
3. If the three sides of an equilateral triangle are produced, all the external acute angles are equal, and all the obtuse angles are equal.
4. If the equal angles of an isosceles triangle are bisected, the triangle formed by the bisectors and the base is an isos- celes triangle.
5. The three straight Imes joining the middle points of the sides of a triangle divide the triangle into four equal tri- angles.
6. If one of the acute angles of a right-angled triangle is double the other, the hypothenuse is double the shortest side.
7. If through any point in the base of an isosceles triangle parallels to the equal sides are drawn, a parallelogram is formed whose perimeter equals the sum of the equal sides of the triangle.
8. If the diagonals of a parallelogram bisect each other at right angles, the figure is either a square or a rhombus.
72 ELEMENTS OF PLANE GEOMETRY.
9. The sum of the four lines drawn to the vertices of any- quadrilateral from any point except the intersection of the diagonals, is greater than the sum of the diagonals.
10. The straight lines which join the middle points of the adjacent sides of any quadrilateral, form a parallelogram whose perimeter is equal to the sum of the diagonals of the given quadrilateral.
11. Lines joining the middle points of the opposite sides of any quadrilateral, bisect each other.
12. If the four angles of a quadrilateral are bisected, the bisectors form a second quadrilateral whose opposite angles are supplements of each other.
Note. — If the figure is a rhombus or a square, there is no second one formed.
ELEMENTS OF PLANE QEOMETBY. 73
PROBLEMS.
1. Show by a diagram that between five points, no three of
5X4 which lie in the same straight line, — ^ — straight lines can
be drawn connecting the points.
2. Between n points, no three of which lie in the same straight line, -^ straight lines can be drawn con- necting the points.
3. What is the greatest number of angles that can be formed with four straight lines ? Ans. 24.
4. If the sum of the interior angles of a ^polygon equals the sum of its exterior angles, how many sides has the polygon?
5. If the sum of the interior angles of a polygon is double the sum of its exterior angles, how many sides has the figure ?
6. If the sum of the exterior angles of a polygon is double the sum of its interior angles, how many sides has the figure ?
74 ELEMENTS OF PLANE GEOMETRY.
BOOK II.
KATio a:n^d propoetiok
DEFINITIONS.
145. To measure a quantity is to find how many times it contains some other quantity of the same kind called the unit of measure.
146. Two quantities are commensurable if they have a common unit of measure.
Two quantities are incommensurable if they have no com- mon unit of measure.
Any two similar quantities may be considered as having a common unit of measure infinitely small.
147. In Geometry we compare two similar quantities by finding how many times one contains the other ; that is, we measure one by the other. The magnitude, therefore, of a quantity is always relative to the magnitude of some other similar quantity.
148. Ratio is the measure of relation between two similar quantities, and is expressed by the quotient resulting from dividing the first by the second.
The first of the two quantities compared is called the Antecedent, and the second the Consequent Taken together they are called the Terms of the Ratio, or a Couplet.
Ratio is indicated by a colon placed between the quantities compared, or by the fractional form of indicating division ;
thus, the ratio of a to 6 is written, a : 6, or -y-
ELEMENTS OF PLANE GEOMETRY, 75
149. A rroportion is an expression of equality between two equal ratios.
Thus, -r = -T. This means that the ratio of a to 6 equals
the ratio of c to d. Usually the proportion is indicated by a double colon placed between the two couplets, thus:
a : b :: c : d.
This is read, a is to b as c is to d; or, the ratio of a to 6 is equal to the ratio of c to d.
Of the four terms compared, the first and third are the Antecedents, and the second and fourth are the Consequents. The Extremes are the first and fourth terms. The Means are the second and third terms. The Fourth Proportional is the fourth term. When the means are equal, as in
a \ b '.'. b '. c,
b is said to be the Mean Proportional between a and c ; and c is said to be the Third Proportional to a and b.
150. Four quantities are Reciprocally Proportional when the first is to the second as the reciprocal of the third is to the reciprocal of the fourth.
Thus, a : 6 ::—:-, •
c a
Two quantities and their reciprocals form a reciprocal proportional.
Thus, a : b :: t- : —
6 a
151. A proportion is takenby JLftema^ion, when antecedent is compared with antecedent, and consequent with consequent.
Thus, if a : 6 : : c : 6?, we have by alternation either a I G :: b i d; ov, d : b :: c : a.
76 ELEMENTS OF PLANE GEOMETRY.
152. A proportion is taken by Inversion^ when the ante- cedents are made consequents and the consequents antece- dents.
Thus, if a : 6 : : c : cf , we have by inversion h : a : : d : c.
153. A proportion is taken by Composition, when the sum of antecedent and consequent is compared with either ante- cedent or consequent.
Thus, if a : h :: c : c^, we have by composition
a -\- h : a :: c -{- d : c;
or a -\- h '. b '.'. c -\- d '. d.
154. A proportion is taken by Dimsion, when the difference of antecedent and consequent is compared with either ante- cedent or consequent.
Thus, if a : h :: c : c^, we have by division
a — h : a :: c — d : c;
or a — h : h '.'. c — did.
155. A Continued rroportion is a series of equal ratios.
Thus, \i a '. h :: h :. c :: c : d :: d : e, we have a continued proportion.
ELEMENTS OF PLANE GEOMETRY. 77
THEOREM I.
156. In any proportion the product of the extremes is equal to the product of the means.
Let a : b :: G : d.
To prove that a d = b c.
ft /»
Take the form r = -r-
Multiply both members by b d; then a d ==b c. Q. E. D.
THEOREM II.
157. If the product of two quantities equals the product of two others, the quantities in either product can be made the means, and those of the other product the extremes of a propor- tion.
Let a d = b c.
To prove that a : b :: c : d. •
Divide both members by b d.
Then 4=4.
b d
or a : b :: c : d. Q. E. D.
7*
78 ELEMENTS OF PLANE GEOMETRY.
THEOREM III.
158. A mean proportional between two quantities equals the square root of their product.
Let a : h :: h : c.
To prove that h = \/ a c, ^
b' = ac. (156)
Extract the square root of both members.
Then b = ^/Ta Q. E. D.
THEOREM IV.
159. The corresponding members of two equations form the couplets of a proportion.
|
Let |
a = c, |
|
and |
b=:d. |
|
To prove that a : |
b :: e: d. |
|
Divide. |
|
|
Then |
a c |
or a : b :: c : d. Q. E. D.
ELEMENTS OF PLANE GEOMETRY. 79
THEOREM V.
160. If four quantities are in proportion, they are in pro- portion hy alternation.
Let a \ b :: c : d.
To prove that a : c :: b : d.
Take the form -— = —-.
6 d
Multiply both members by —
Then ^=A,
c a
or a : c :: b : d. Q.E.D.
THEOREM VI.
161. If four quantities are in proportion^ they are in pro- portion by inversion.
Let a : b w c '. d.
To prove that b : a :: d : e.
Take the form — = —-.
0 a
Divide 1 by each member.
Then ^ = 1,
a e
or b : a :: d : c, Q.E.D.
80 ELEMENTS OF PLANE GEOMETRY.
THEOREM VII.
162. If Jour quantities are in proportion^ they are in pro- portion by composition.
Let a : b :: c \ d.
To prove that a -\- b : b :: c -\- d : d.
Take the form t~ = ~r*
b a
To each member add 1.
Then |.4-i:^-|.4_l;
whence
a -\- b c -{- d
b d
or a^b -.b :: G-\- d \ d. Q. E. D.
THEOREM VIII.
163. If Jour quantities are in proportion, they are in pro- portion by division.
Let a : b w G : d.
To prove that a — b : b :: c — d : d.
|
Take the form |
a |
c ~d' |
||
|
From each member subtract 1. |
||||
|
Then |
a T |
— 1 = |
c |
-1; |
|
bence |
a — b |
_6 _c |
— d d ' |
|
|
a — b : |
b :: c- |
-d : |
d. |
or a — b : b :: c — d : d. Q.E.D.
ELEMENTS OF PLANE GEOMETRY. 81
THEOREM IX.
164. If two proportions have a couplet in each the same, the other couplets form a proportion.
|
Let a : b :: G : d, |
||
|
and a : b :: e : f. |
||
|
To prove that c : d :: e : f. |
||
|
Take the forms -- = -5- and -y- = 0 a 0 |
e 7* |
|
|
Then 4 = 4-» |
(Ax. 1.) |
|
|
or c : d :: e : f. |
Q.E.D. |
THEOREM X.
165. Equimultiples of two quantities are proportional to the quantities themselves.
Let a and b be any two quantities.
To prove that ma : mb :: a : b.
a a
Multiply both terms of the first member by m.
Then '^ = -%,
mb 0
or ma : mb :: a : h Q. E. D.
82 ELEMENTS OF PLANE GEOMETRY.
THEOREM XI.
166. In any proportiony any equimultiples of the first couplet are proportional to any equimultiples of the second couplet.
Let a : b :: c : d.
To prove that ma : mb :: nc : nd.
Take the form -y- = ^r*
0 a
Multiply both terms of the first member by m, and both terms of the second member by n.
Then
ma 7ic
mb nd
or
ma : mb :: nc : 7id. Q. E. D.
THEOREM XII.
167. If two quantities are increased or diminished by like parts of each, the results are proportional to the quantities themselves.
Let a and b be any two quantities.
To prove that a ±. — a : b ± — b :: a : b.
ma : mb :: a : b. (165)
For m, substitute 1 ±: — •
Then (l ± ^^ a : (l it ^) b :: a : b,
q/ \ q
P-a:b±P-
or a± ^ a : b ±^ b :: a : b. Q. E. D.
ELEMENTS OF PLANE GEOMETRY. 83
THEOREM XIII.
16.8. In any continued proportion, the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent.
Let a : h :'. c : d w e : f '.: g : h.
To prove that a -\- c -\- e -\- g : b -{- d -{- f -\- h :: a : b.
Denote the common ratio by r.
mi a c e a
Then r=-^=^^ = j = JL.
Whence a = br, c = dr, e = fr, g = hr. Add these equals.
Then a + c + e + ^ = (64-^+/+/i)r. Divide both members hy (b-\-d-\-f-^h). rru^^ a + c-j- e + g a
b -{- d +/+/1 b
or a 4- c + e + ^ : 6 + ^ +/ -f /i :: a : 6. Q. E. D,
THEOREM XIV.
169. In two or more proportions, the products of the corre- sponding terms are proportional.
C a : b \: c : d,
Let <^ e : f :: g : h, ^
\m '. n w 0 : p.
To prove that aem : bfn : : ego : dhp.
Take the forms -^ = — r? --r = 4-^ — = — 0 d J h n p
By multiplication we have
aem ego
bfn dhp or aem : bfn : : ego : dhp. Q. E. D.
84 ELEMENTS OF PLANE GEOMETRY.
THEOREM XV.
170. Like powers, or like roots, of the terms of a proportion form a proportion.
Let a : h :: c : d.
To prove that aJ" : 6" :: c" : d", and that a" : 6" : : c" : d~.
Take the form -7- = — 7-
0 a
Raise each member to the ?i'* power.
Then 4l = 4"'
V d"
or a" : 6" :: c" : c?".
Also extract the ?i'* root of each member. '
Then — == — '
b^ d-^
or a" : 6" : : c" : d». Q. E. D.
171. Scholium. The product of two quantities implies that at least one is numerical.
In (169) and (170), all the quantities must be numerical.
In (160) and (168), all the quantities must be of the same kind.
ELEMENTS OF PLANE GEOMETRY. 85
BOOK III.
THE CIRCLE.
DEFINITIONS.
172. A Circle is a plane bounded by a curve, all the points of which are equally distant from q^ a point within, called the centre.
A\ The Circumference of a circle is the curve which bounds it.
An Arc is a part of the circumference ; as, ^ C. A Semi- circumference is an arc equal to half of the circumference.
A Radius is a straight line extending from the centre to any point in the circumference; as, OC.
173. A Diameter of a circle is a straight line passing through the centre and terminating each way in the circum- ference; as, AB.
174. A Chord is a straight line joining any two points in the circumference; as, P/
^^' JD^ '0
The arc EPD is said to be subtended by its chord ED. Every chord subtends two arcs, whose sum equals the whole circumference. Whenever an arc and its chord are spoken of, the less arc is meant.
8
8Q
ELEMENTS OF PLANE GEOMETRY.
' 175. A Segment of a circle is the portion enclosed by an arc and its chord; as, FGH. A Semicircle is a segment equal to one-half of the circle.
176. A Sector of a circle is the por- tion enclosed by an arc and the radii drawn to its extremities ; as, OFM.
111. A Tangent is a straight line which touches the circumference but does not intersect it; as, ABC.
The common point B is called the point of contact, or the point of tangency.
178. A Secant is a straight line which cuts the circumference in two points ; as, DE.
179. An Inscribed Angle is one whose vertex is in the circumference and whose sides are chords; as, ABE.
180. An Inscribed, Polygon . is one
whose sides are chords of a circle; as, B^ D
ABCDEF. The circle is then said to be circurtiscrihed about the polygon.
181. A Polygon is circumscribed about a circle when all its sides are tangents to the circumference; as, MNOPQ. The circle is then said to be inscribed.
182. By the definition of a circle, all its radii are equal; also, all its Q diameters are equal. It also follows from the definition that circles are equal when their radii are equal.
ELEMENTS OF PLANE GEOMETRY. 87
CHOEDS, ARCS, AISTGLES AT THE CENTRE, SECANTS, AND RADII.
THEOREM I.
183. Any diameter bisects the circle and its drcumference. Let ABCD be a O, and AB any diameter.
C
D
To prove that ABC= ABD.
On AB as an axis, revolve the portion ABC till it falls in the plane of -4-Si).
Then the curve ACB coincides with the curve ADB, for all the points in each are equally distant from the centre 0.
ABC = ABD, (14)
and curve ACB = curve ADB. Q. E. D.
88 ELEMENTS OF PLANE GEOMETRY.
THEOREM II.
184. A diameter of a circle is greater than any other chord.
In the O ACB, let AB he any diameter and BC any other chord.
ToprovethatAB> BC.
From the centre 0, draw OC, '
0A= Oa (182)
0C-\- 0B> BC. (Ax. 11)
Substitute OA for its equal OC. Then 0A-\- 0B> BC,
or AB>Ba (Q.E.D.)
ELEMENTS OF PLANE GEOMETRY.
89
THEOREM III.
185. A straight line cannot cut the circumference of a circle in more than two points.
In the O ADBC, let AB cut the circumference at A and B.
To prove that A B cannot cut the circumference in more than two points.
Draw the radii OA, OB, OD,
0A= 0B= OD.
(182)
li AB could cut the circumference at A, B, and 2), there would be three equal straight lines drawn from the same point to the same straight line, which cannot be. (61)
. • . a straight line cannot cut a circumference in more than two points. Q. E. D.
8*
90
ELEMENTS OF PLANE GEOMETRY.
THEOREM IV.
186. In equal circles, or in the same circle, equal angles at the centre iyitercept equal arcs.
Let 0 and P be the centres of the equal Os ABC and DEF, and let Z. a = ^ 5.
To prove that arc AB = arc DE.
Place the OABC on the ODEF so that Z_ a coincides with Z_ h.
0A = OB = PD = PE; (182) A falls on D, and B falls on E;
ABO = DEP, (14)
and arc AB = arc DE. Q. E. D.
187. Cor. — Li equal circles, or in the sayne circle, equal arcs subtend equal angles at the centre.
ELEMENTS OF PLANE GEOMETRY.
91
THEOREM V.
188. In equal circles, or in the same circle, equal arcs are subtended by equal chords.
In the equal Os ABE and CDF, let arc AB = arc CD,
E F
To prove that chord AB = chord CD. Draw the radii OA, OB, PC, PD.
OA = PC, (182)
OB = PD, (182)
and Z.a=/-b; (187)
AABO = ACDP, (83)
and chord AB = chord CD. Q. E. D.
189. Cor. — In equal circles, or in the same circle, equal chords subtend equal arcs.
92
ELEMENTS OF PLANE GEOMETRY,
THEOREM VI.
190. TJie radius petpeyidicular to a chord bisects the chord and the subtended arc.
Let OCbe a radius _L to the chord AB at D.
To prove that AD = BD, and arc AC = arc BC. Draw the radii OA and OB.
and
and Also
0A= OB, (182)
OD = OD; RA AOD = KA BOD, (91)
AD = BD. Z. a= ^ b; 2irG AC = fiYG BC. (186) Q. E. D.
191. Cor. — The perpendicular erected at the middle point of a chord passes through the centre of the circle and bisects the subtended arc.
ELEMENTS OF PLANE GEOMETRY. 93
THEOREM VII.
192. In the same circle, or in equal circles, equal chords are equally distant from the centre; and if two chords are unequal, the less is at the greater distance from the centre.
Let chord AB = chord CD, and chord CE < chord CD; and let OF, 0 G, and OH be _Ls to these chords from the centre 0.
To prove that OF=OG, and OH > OG.
OF and OG bisect the equal chords ^5 and CD; (190) AF= CG; 'RAAOF='RA COG, (91)
and 0F= OG.
Again, CD > CE;
OH cuts CD in some point, as P. Now 0H> OP.
But 0P> OG; (62)
still more is OH > 0 G. Q. E. D.
94 ELEMENTS OF PLANE GEOMETRY.
THEOREM VIH.
193. Through three points not in the same straight line, a circumference of a circle can be passed.
Let A, B, and C be any three points not in the same straight line.
To prove that through A, B, and C, a circumference of a O can he passed.
Draw AB and BC, and at their middle points let the -Ls EM and DP be erected.
AB and B C are not in the same straight line ;
the _J_s EM and DP meet in some point, as O.
Now, 0 is equally distant from A and B; also from B and C; (53)
. • . O is equally distant from A, B, and C, and a circum- ference with OA as a radius passes through these points.
Q.E.D.
ELEMENTS OF PLANE GEOMETRY.
95
THEOREM rx,, .e- **X]
194. A straight line perpendicular to a radimlit^ extrehiity is a tangent to the circle.
Let AB be J_ to the radius OP at P.
To prove that AB is a tangent to the O a^ the point P.
From the centre draw any oblique line, as OG
0C> OP; (52)
. • . the point C is without the circumference, and all points in AB, except P, are without the circumference;
AB is a tangent to the O at P. (177) Q. E. D.
195. Cor. — A straight line tangent to a circle is perpen- dicular to the radius drawn to the point of contact.
ELEMENTS OF PLANE GEOMETRY.
THEOREM X.
196. Two parallel secants intercept equal arcs.
Let the lis AB and CD intercept the arcs ^C and BD.
To prove that arc AC = arc BD.
Suppose the radius OE to be drawn _L to AB and CD.
Then arc AE = arc BE,
and &TC CE= arc DE. (190)
Subtract; then arc AE — arc CE = arc BE — arc DE.
or arc AC = arc BD. Q. E. D.
197. Scholium. — This proposition is true for any position of tjie parallels; hence it is true if one or both become tangents ; and the straight line which joins the points of contact of two parallel tangents is a diameter.
ELEMENTS OF PLANE GEOMETRY. 97
EELATIYE POSITION OF CIECLES.
THEOREM XI.
198. If two circles cut each other, the straight line joining their centres bisects their common chord at right angles.
Let AB be a common chord of two O s which cut each other, and 00 join the centres 0 and O.
To prove that 0 C is -L to AB at its middle point. Draw the radii OA, OB, CA, and CB.
OA = OB, (182)
and CA = CB; (182)
0 is equally distant from A and B, and Ois equally distant from A and B;
OCis J- to AB eit its middle point. (55) Q. E. D.
199. Cor. — If two circles touch each other, either externally or internally, the point of contact is in the line joining their centres.
98 ELEMENTS OF PLANE GEOMETRY.
THEOREM XII.
200. If two circles cut each othevy the distance between their centres is less than the sum and greater than the difference of their radii.
Let 0 and C be the centres of two Os whose circumferences cut each other at A and B, and draw the radii OA and CA.
To prove that 0C< OA + CA, and 0C> OA — CA. The point A does not lie in OC; (199)
OC^ is a A. Now, OC<OA-\- CA, (Ax. 11)
and 0C> OA— CA. (73) Q. E. D.
201. Cor. 1. — If two circles touch each other externally, the distance between their centres equals the sum of their radii.
202. "Cor. 2. — If two circles are wholly exterior to each other, the distance between their centres is greater than the sum of their radii.
ELEMENTS OF PLANE GEOMETRY.
99
THE MEASUEEMEE-T OF AITGLES.
THEOREM XIII.
203. In equal circles, or in the same circle, angles at the centre are to each other as the arcs which they intercept.
In the equal Os ABM and CDN, let 0 and P be the centres, and let the Z.s AOB and CPD intercept the arcs AB and CD.
M N
To prove that A. AOB : Z. CPD :: arc AB : arc CD.
Let EF be a common unit of measure of AB and CD, and suppose it to be contained in AB 8 times, and in CD 5 times.
Then arc AB : arc CD :: S : 5.
Draw radii at the several points of division of the arcs.
Then the partial ZL s are equal. (187)
AOB contains 8, and CPD contains 5 equal Z-s;
/LAOB : Z. CPD :: 8 : 5.
But arc^-B : arc CD :: S : 5;
/. AOB : Z. CPD :: arc AB : arc CD.
100 ELEMENTS OF PLANE GEOMETRY.
The same proportion is found if other numbers than 8 and 5 are taken.
Now, this is true, whatever may be the length of the com- mon unit of measure of the arcs; hence it is true when it is iniinitely small, as is the case when the arcs are incom- mensurable. Hence, in any case, the proposition is true.
Q.E.D.
204. Cor. — In equal circles^ or in the same circle^ arcs which are intercepted by angles at the centre are to each other as the angles.
205. Scholium. — The truth of this proposition gave rise to the method of measuring angles by arcs. It will be observed, that if arcs are struck with the same radius from the vertices of angles as centres, the angles are to each other as the arcs intercepted by their sides. Hence the angle is said to be measured by the arc.
The unit of measure generally adopted is an arc equal to 3 J (J of the circumference of a circle, called a degree, and denoted thus (°).
The degree is divided into 60 equal parts, called minutes, denoted thus, (').
The minute is divided into 60 equal parts, called seconds, denoted thus, (").
A right angle, therefore, is measured by 90°; or, as we say, it is an angle of 90°.
An angle of 45° is ^ of a right angle ; an angle of 30° is i of a right angle. Thus we have a definite idea of the magnitude of an angle if we know the number of degrees by which it is measured.
ELEMENTS OF PLANE GEOMETRY.
101
THEOREM XIV.
206. An inscribed angle is measured by half the intercepted arc.
In theO^-BC, let AB and CB be the sides of the in- scribed Z. a.
Case I. — Let the centre 0 be in one of the sides.
B
|
To prove that Z_ a is measured by i arc A C. |
||
|
Draw the radius OA. |
||
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OA = OB; |
(182) |
|
|
. |
^ ABO is isosceles, |
|
|
id |
z:a = Z.b. |
(94) |
|
Z-c = a /- -^ L.b. |
(82) |
|
|
Substitute |
Z. a for its equal Z. b. |
|
|
Then |
Z. c = 2 21 a. . - |
|
|
Now, id |
Z. c is measured by arc A C; 2 Z_ a is measured by arc A C, Z_ a is measured by 2 A C. 9* |
(205) |
102
ELEMENTS OF PLANE GEOMETRY.
Case 11.— Let the centre 0 fall within the inscribed Z_.
B
To prove that Z_ ABC is measured by ^ arc A C. Draw the diameter BD.
Z. a is measured by ^ arc AD, and Z. 6 is measured by ^ arc D C; (Case I.)
Z^ a -\- /- b\^ measured by ^ (arc AD + arc DC), or Z. ABC is measured by h AC.
Case III. — Let the centre 0 fall without the inscribed Z_ .
B
To prove that Z_ a is measured by i arc AC.
Draw the diameter BD.
Z. DBC is measured by ^ arc DC, and Z. 6 is measured by i arc DA ; (Case I.)
. • . Z. D^C — Z_ 6 is measured by J (arc DC — arc DA), or Z. a is measured by ^ arc J. C Q. E. D,
ELEMENTS OF PLANE GEOMETRY. 103
.^ ft .
207. Cor. 1. — All angles inscribed in the same segment are equal.
208. Cor. 2. — Any angle inscribed in a semicircle is a right angle.
209. Cor. 3. — Any angle inscribed in a segment less than a semicircle is obtuse.
210. Cor. 4. — Any angle inscribed in a segment greater than a semicircle is acute.
104 ELEMENTS OF PLANE GEOMETRY.
THEOREM XV.
211. Any angle formed by a tangent and a chord is measured by half the intercepted arc.
Let the Z. ABD be formed by the tangent CD and the chord AB.
To prove that Z. ABD is measured by J arc AEB.
Draw the diameter BE. Z. 6 = a L, (195) and is measured by ^ arc EB, (208) and Z. a is measured by h arc AE; (206)
. • . Z- a + Z_ 6 is measured by h (arc AE -{- arc EB), or Z. ABD is measured by J arc AEB. Q. E. D.
ELEMENTS OF PLANE GEOMETRY. 105
THEOREM XVI.
212. Any angle formed by two chords intersecting is measured by half the sum of the arcs intercepted between its sides and the sides of its vertical angle.
Let the Z. a be formed by the chords AB and CD.
To prove that Z. a is measured by ^ (arc AD -\- arc GB).
DrawylC.
Z. c is measured by h arc AD, . and Z. 6 is measured by l arc CB. (206)
But ^a= /_ c^ l.h; (82)
Z. a is measured by h (arc AD + arc GB).
Q.KD.
106
ELEMENTS OF PLANE GEOMETRY.
THEOREM XVH.
213. Any angle formed by two secants is measured by half the difference of the intercepted arcs.
Let the Z. a be formed by the secants AB and BC.
B
To prove that /L a is measured by I {arc AC — arc EF). Let Z)£ be II to O^.
arc DC = arc EF. (196)
Z. 6 is measured by i arc AD.
But arc AD = (arc AC — arGDC) = (arc ^ O—- arc EF,)
and Z^b= Z. a; (62)
Z. a is measured by i (arc AC — arc EF). Q. E. D.
214. Scholium. — This proposition is true for any position of the secants; hence it is true if one or both become tangents.
ELEMENTS OF PLANE GEOMETRY. 107
PEOBLEMS IN COIsrSTEUCTIOK
PROBLEM I.
215. To bisect a given straight line.
Let AB be the given straight line.
B
/E\
With A and B as centres and a radius greater than half of AB, describe arcs intersecting at C and E.
Draw CE.
Then CE bisects AB at the point P. (55) Q.E.F.
108
ELEMENTS OF PLANE GEOMETRY,
PROBLEM n.
216. At any point in a straight line to erect a perpendicular. Let P be any point in the straight line AB.
D
E
Cut off PD = PE.
With D and E as centres and a radius greater than PD or PE^ describe arcs intersecting at C
Draw CP.
Then
CPh A- to AB. (55) Q.E.F.
ELEMENTS OF PLANE GEOMETRY. 109
PROBLEM III.
217. From any point without a straight line to draw a per- pendicular to that line.
Let P be any point without the straight line AB.
^ c--^,,.,.|.,.,.^--^
.^^^
With P as a centre and a radius sufficiently great, describe an arc cutting AB 2X C and D.
With C and B as centres and a radius greater than half of CD^ describe arcs intersecting at E.
Draw FE.
Then isPjEJLto^^. (65) Q.E.F.
10
110
ELEMENTS OF PLANE GEOMETRY.
PROBLEM IV.
218. To bisect a given arc. Let AOB be a given arc.
/\
0 X./
♦ Draw the chord AB.
Bisect AB by a _1_ as in (215). This _L bisects the arc.
(191) Q.E.F.
ELEMENTS OF PLANE GEOMETRY. Ill
PROBLEM V.
219. To construct an angle equal to a given angle^ at any point in a line.
Let a be the given /_ , and A the point in the line AB.
V
A^^ L B
With the vertex 0 as a centre, and any radius, describe an arc cutting the sides of Z. a at if and N.
With ^ as a centre and the same radius, describe the indefinite arc CD.
Draw the chord M N.
With D as a centre and iltf iV as a radius, describe an arc cutting the indefinite arc at C.
Drawee.
Then arc CD = arc M N; (189)
/.h=l.a. (187) Q.E.F.
112
ELEMENTS OF fLANE GEOMETRY,
PROBLEM VI.
220. To bisect a given angle. Let ABC hQ the given Z..
With J5 as a centre and any radius, describe an arc cutting the sides of the Z. at i) and E.
Bisect the arc as in (218).
Then, since arc DF = arc FE, Z. a= ^h. (187)
Q.E.F.
PROBLEM VH.
221. Through a given point, to draw a straight line parallel to a given straight line.
Let P be the given point, and AB the given straight line.
D
PA
/
/a
/b
From any point in AB, as C, draw the line CD through P.
At P construct the Z- fe = Z. a as in (219). Then P^ is II to ^5. (64) Q.E.F.
ELEMENTS OF PLANE GEOMETRY. 113
PROBLEM VIII. 222. CH^ven two angles of a triangle, to find the third angle. Let a and h be the given Z.8.
./
A -CL-lw
P Draw the indefinite line A B.
At any point in A B, as P, construct
/- m = Z- a, and Z. n = Z. 5, as in (219).
Then Z. c is the third Z. of the A. (77) Q. E. F.
PROBLEM IX.
223. 7\vo sides and the included angle of a triangle being given, to construct the triangle.
Let m and n be the given sides, and a the included Z. .
E
|
A_ |
. A |
K |
|
|
m |
|||
|
n |
\ |
||
|
Draw the indefinite line AB. |
C |
||
|
Cutoff |
AC = |
= n. |
|
|
At A construct Z_ h -- |
= Z^a. |
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On AE Q\xt oS AD = |
m. |
||
|
Draw CD. |
Then A CD is the required A. (83) Q. E. F.
10*
114 ELEMENTS OF PLANE GEOMETRY.
PROBLEM X.
224. One side and two adjacent angles of a triangle being given, to construct the triangle.
Let AB he the given side, and a and b the given Z_g.
\V
/\
/ \
At A construct Z. c = Z^a, and at B construct Z- d = Z-b.
The sides AD and BE intersect at (7, '
and ABC is the required A. (84) Q. E. F.
225. Scholium. This problem is not possible if the sum of the two given angles is equal to or greater than two right angles.
ELEMENTS OF PLANE GEOMETRY. 115
PROBLEM XI.
226. Given the three sides of a triangle, to construct the triangle.
|
Let m, n, and o |
be the three sides of a A. |
||
|
n |
A |
||
|
m |
's. |
||
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0 |
\ |
||
|
Z A |
\ |
||
|
D 1 |
|||
|
Draw an indefinite line AB. |
|||
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It off AD = 0. |
With ^ as a centre and m as a radius, describe an arc; and with D as a centre and n as a radius, describe an arc cutting the first at C.
Drawee and DG
Then ADC is the required A. (86) Q. E. P.
227. Scholium. — This problem is not possible if one side is equal to or greater than the sum of the other two sides.
116
ELEMENTS OF PLANE GEOMETRY.
PROBLEM XII.
228. Gfiven two sides of a triangle and the angle opposite one of them, to cdnstrud the triangle.
Let m and n be the given sides, and a the given Z. opposite the shorter side m.
I. When the given angle is acute, and the side opposite is less than the other given side.
D
|
n |
A. |
|
|
Xa |
y' |
|
|
::;*^^ |
.r
...v ^
Construct /^ h = /L a, and on AD cut o^ AC = n.
With (7 as a centre and m as a radius, describe an arc cutting the side AE at B and P.
Draw CB and CP.
Then either ABC or APC is the required A, and there are two solutions to the problem.
When m equals the J_ CF, there is but one construction.
When m is less than CF, the problem is not possible.
ELEMENTS OF PLANE GEOMETRY. 117
II. When the given angle is acute, right, or obtuse, and the side opposite is greater than the other given side.
E
|
m |
Cy^' |
|
|
n |
||
|
X |
Jv:</< ^,/._. |
F |
|
y^ d |
d\A /B |
When the given Z_ a is acute, construct Z^ b = Z^ a, and cut off on the side AE, AC = n.
With (7 as a centre and m, the greater side, as a radius, describe an arc cutting the side AF at B, and AF produced atZ).
Draw CB and CD.
Then ABC is the required A; and there is but one solution. Q. E. F.
When the given Z. is obtuse, as c, the A A CD is the required one ; and there is but one solution.
When the given Z. is a L, the problem has two solutions. Let the pupil give the construction.
229. Scholium. — The problem is not possible if the given angle is right or obtuse, and the side opposite is less than the other given side.
118 ELEMENTS OF PLANE GEOMETRY. .
PROBLEM Xin.
230. Given two sides and the included angle of a parallelo- gram, to construct the parallelogram.
Let m and n be the given sides, and a the included Z. .
Construct /Lh = Z^ a, and on the sides cut off J. C and AB respectively equal to m and n.
With ^ as a centre and J.C as a radius, describe an arc; with C as a centre and AB as a radius, describe an arc cutting the other at D.
Draw BD and CD.
Then ABCD is the required EJ. (132) Q.E.F.
ELEMENTS OF PLANE GEOMETRY. 119
PROBLEM XIV.
231. To find the centre of a given drde, Jjet ABC D be a given O-
From any point in the circumference, as B, draw two chords, AB and BC.
Bisect AB and BC hj -Ls as in (215).
The point 0, the intersection of the JLb, is the required centre. (191) Q. E. F.
120
ELEMENTS OF PLANE GEOMETRY,
PROBLEM XV.
232. At a given point in the circumference of a circle^ to draw a tangent to the circle.
Let P be the given point in the circumference of the O EDP.
If the centre is not given, find it by (231).
Draw the radius OP, and at P draw AB _L to OP.
Then AB is the required tangent.
(194) Q.E.F.
ELEMENTS OF PLANE GEOMETRY. 121
PROBLEM XVI.
233. TJirough a given point without a given circle, to draw a tangent to the circle.
Let 0 be the centre of the given O , and P the given point.
:MP
Draw OP, and upon it describe a circumference cutting the given circumference at A and C.
Draw AP and CP; also the radii OA and OC.
/_OAP=rL', (208)
AP is _L to OA, and is tangent to the O- (194)
Likewise we can prove that CP is a tangent. Q. E. F.
234. Cor. — From any point without a given circle, two equal ta7igents to the circle can be drawn.
11
122
ELEMENTS OF PLANE GEOMETRY.
PROBLEM XVII.
235. To inscribe a circle in a given triangle. Let ABC hQ the given A.
Bisect i\\Q/-sABC and CAB.
The bisectors meet in some point, as 0. '
From 0 draw OF.OD, and OE, J- to the sides of the A.
The J_3 are all equal. (104)
With 0 as a centre and OF as a radius, describe a circle.
Then FED is the required O. (181) Q. E. F.
ELEMENTS OF PLANE GEOMETRY.
123
PROBLEM XVIII.
236. On a given straight line^ to describe a segment of a circle which shall coyitain a given angle.
Let AB be the given line, and a the given Z. . E
/ / \ N
/F \ \
A^-
/^-.
\ /
0 \
D
'^YB
At B construct the Z. AB C =^ /- a.
Draw(^^_Lto jBCat^.
Bisect AB, and at its middle point erect the J_ BF.
With 0, the intersection of i)i^ and G^J5, as a centre and 0^ as radius, describe a circumference.
Now, BC is J_ to the radius OB;
BC is R tangent to the circle. (1^4)
The Z.ABC is measured by * arc AB. (211)
But Z. b, the Z. inscribed in the segment J. ^^, is measured by i arc AB ; (206)
AEB is the required segment.
Q.E.F.
124 ELEMENTS OE PLANE GEOMETRY.
EXERCISES 1^ INYENTIOK
THEOREMS.
1. In any circumscribed quadrilateral, the sura of two opposite sides is equal to the sum of the other two sides.
2. A quadrilateral is inscriptible if two of its opposite angles are supplements of each other.
3. The bisectors of the angles formed by producing the opposite sides of an inscribed quadrilateral intersect at right angles.
4. If a circle is described on the radius of another circle, any straight line drawn from the point of contact to the outer circumference is bisected by the interior one.
PROBLEMS.
1. To trisect a right angle.
2. Given two lines that would meet if sufficiently pro- duced, to find the bisector of their included angle without finding its vertex.
3. To draw a common tangent to two given circles.
4. Inscribe a square in a given rhombus.
5. To construct a square, given its diagonal.
6. Construct an angle of 30°, one of 60°, one of 120°, one of 150°, one of 45°, and one of 135°.
7. Construct a triangle, given the base, the angle opposite the base, and the medial line to the base.
ELEMENTS OF PLANE GEOMETRY. 125
8. Construct a triangle, given the vertical angle, and the radius of the circumscribing circle.
9. Construct a triangle, given the base, the vertical angle, and the perpendicular from the extremity of the base to the opposite side.
10. Construct a triangle, given the base, an angle at the base, and the sum or difference of the other two sides.
11. Construct a square, given the sura or difference of its diagonal and side.
12. Describe a circle cutting the sides of a square, so as to divide the circumference at the points of intersection into eight equal arcs.
13. Through any point within a circle, except the centre, to draw a chord which shall be bisected at that point.
11*
126 ELEMENTS OF PLANE GEOMETRY.
BOOK IV.
AEEA A1^T> RELATION OF POLYGONS. .
DEFINITIONS.
237. Similar Polygons are polygons which are mutually equiangular, and have their homologous sides proportional.
238. The Area of a polygon is its quantity of surface; it is expressed by the number of times the polygon contains some other area taken as a unit of measure. The unit of measure usually assumed is a square, a side of which is some linear unit; as, a square inch, a square foot, etc.
239. Equivalent Figures are such as have equal areas.
ELEMENTS OF PLANE GEOMETRY.
127
AREAS.
THEOREM I.
240. Tlie area of a rectangle equals the product of its base and altitude.
Let ABCD be a rectangle, AB the base, and ^(7 the altitude.
D
B
To prove that the area of ABCD ^ AB X AC.
Let AE be a common unit of measure of the sides AB and A C, and suppose it to be contained in AB 5 times, and in J.C 3 times.
Apply AE to AB and AC, dividmg them respectively into five and three equal parts.
Through the several points of division draw -Ls to the sides.
The rectangle will then be divided into equal squares, as the angles are all Ls, and the sides all equal. (130)
128 ELEMENTS OF PLANE GEOMETRY.
Now, the whole number of these squares is equal to the number in the row on AB multiplied by the number of rows, or the number of linear units in AB multiplied by the number in J.C.
Now, this is true, whatever may be the length of the common unit of measure; hence it is true if it is infinitely small, as is the case when the sides are incommensurable. Therefore, in any case, the proposition is true. Q. E. D.
ELEMENTS OF PLANE GEOMETRY.
129
THEOREM II.
241. Rectangles are to each other as the products of their bases and altitudes.
Let R and r denote the areas of the rectangles whose bases are B and b, and whose altitudes are A and a.
To prove that R : r :: A X B : a X b, R = AXB, and r= a X b; (240)
R : r :: A X B : a X b. (159) Q.E.D.
242. Cor. — Rectangles having equal bases are to each other as their altitudes; rectangles haviiig equal altitudes are to each other as their bases.
130 ELEMENTS OF PLANE GEOMETRY.
THEOREM III.
243. The area of a parallelogram equals the product of its base and altitude.
Let ABEC be a ZZ7, AB its base, and CD its altitude.
0 E
D B
To prove that the area of CD ABEC == AB X CD.
Construct the rectangle DFEC having the same base and altitude as the CJ.
AC = BE, and DC =FE; RAACD = IIA BEF. (91)
Remove A A CD, and rectangle DFEC remains. Remove A BEF, and O ABEC remains; rectangle DFEC = CO ABEC. But the area of the rectangle DFEC = AB X CD;
the area of the o ABEC = AB X CD. Q. E. D.
244. Cor. 1. — Parallelograms are to each other as the pro- ducts of their bases and altitudes.
245. Cor. 2. — Parallelograms having equal bases are to each other as their altitudes; parallelograms having equal altitudes are to each other as their bases.
246. Cor. 3. — Parallelograms having equal bases and equal altitudes are equivalent figures.
ELEMENTS OF PLANE GEOMETRY. 131
THEOREM IV.
247. The area of a triangle equals half the product of its base and altitude.
Let ABC be a A, AB its base, CD its altitude.
B ^
To prove that the area of the A ABC = * {AB X CD).
Through C draw CE II to AB, and through A draw AE II to BC.
The A ABC 'is half of the o ABCE; (129)
but the area of the CJ ABCE == AB X CD; (243)
the area of the A ABC = i (AB X CD). Q. E. D.
248. Cor. 1. — Triangles are to each other as the products of their bases and altitudes.
249. Cor. 2. — Triangles having equal bases are to each other as their altitudes ; triangles having equal altitudes are to each other as their bases.
250. Cor. 3. — Triangles having equal bases and equal alii' tudes are equivalent figures.
132
ELEMENTS OF PLANE GEOMETRY.
THEOREM V.
251. The area of a trapezoid is equal to half the sum of its parallel sides multiplied by its altitude.
Let ABJJD be a trapezoid, AB and DC its II sides, and EF its altitude.
^^-k
"■^B
To prove that the area of ABCD = ] {AB -^ DC) EF,
Draw the diagonal BD, forming As ABD and CDB.
1(he area of the A ABD = I AB X EF,
and the area of the A CDB =1 DC X EF; (247)
.'.l^ABD^l\CDB=^2iVQ?.oiABCD=l{AB-\-DC)EF.
Q. E. D.
252. CoR.--27ie area of a trapezoid is equal to the product of the line joining the middle points of its non-parallel sides and its altitude.
253. Scholium. — The area of an irregular polygon is found by finding the areas of the several triangles into which it can be divided. In survey- ing, the method usually resorted to is to draw the longest diagonal, and to draw perpen- diculars to this diagonal from the other vertices, thus dividing the polygon into rectangles, right-angled triangles, and trapezoids. The areas of these figures are then readily found.
ELEMENTS OF PLANE GEOMETRY.
133
SQUARES O^ LIIsTES.
THEOREM VI.
254. Tlie square described on the hypothenuse of a right- angled triangle is equivalent to the sum of the squares on the other two sides.
Let ABC be a KA, right-angled at C. G
H
\cy
M
To prove that AB = AC -]- BG
On AB, AC, and BC, construct the squares AE, AG, and BH.
Through C draw'C^ II to BE, and draw ^i^and CD. /^ ACB = SiL (Hyp.), and Z. a ::=^ a L ; (Cons.) GCB is a straight line.
12
134 ELEMENTS OF PLANE GEOMETRY.
For a like reason A CH is a straight line,
AB =.. AD, and AF = A C, (Cons.)
and Z_ FAB = Z. CAD, each being a L ~\- -^ b;
.-. . A J.J5i'^= A ^CD. (83)
Square AG and A ABF have a common base J.i^ and a common altitude ;
. • . Square A G is double A ABF; (240) and (247)
and for a like reason
/U AK is double A A CD.
But A A CD = A ABF;
ZZ7 AK = square A G.
Likewise we can prove ZZ7 BK = square BH;
.-. cj AK-^ OJ BK=AB'' = AZ'' X BC\ (Q.E.D.)
255. Cor. 1. — TJie square on either side about a right-angled triangle is equivalent to the square on the hypothenuse minus the square on the other side.
256. Cor. 2. — The square on the diagonal of a square is double the given square. i
257. Cor. Z. — The diagonal and the side of a square are
incommensurable.
Let d be the diagonal, and a the side of a square.
Then d^ = a' -\- a' =^ 2 a\
Extract the square root of each member.
Then d — a \/'2.
Divide by a;
d ,_
then — = i/2 = 1. 41421 +.
ELEMENTS OF PLANE GEOMETRY. 135
PROJECTIOK
DEFINITIONS.
258. The Projection of a Point upon an indefinite straight line is the foot of the perpendicular drawn from the point to the line.
Thus, the projection of the point C upon the line AB is the point E.
C
E
The Projection, of a Finite Straiffht Line upon an in- definite one, is the part of the line intercepted between the perpendiculars drawn from the extremities of the finite line. Thus, EF is the projection of CD upon AB.
If one extremity of CD rests upon the other line AB, then the projection of CD is ED.
136
ELEMENTS OF PLANE GEOMETRY.
THEOREM VII.
259. In any triangle, the square on the side opposite an acute angle equals the sum of the squares of the other two sides minus twice the product of one of those sides and the projection of the other upon that side.
In the A ABC, let c be an acute /_, and PC the pro- jection of AC upon BC.
To prove that AB^ =r BC' + ^C' — 2 BC X PC.
If P falls on the base,
PB = BC—Pa If P falls on the base produced,
PB = PC—BC. In either case, by Algebra,
PB' := BC' + PC' — 2BCX PC,
Add AP to each member; then AP' + PB' = BC" -\- AP' -\-PC' — 2BCX PC.
But AB' = AP' + PB\
and AXf ^AP' -{- PC'- (254)
Substitute AB and AC for their equals; then AB' = BC' -f AC' — 2BCX PC. Q. E. D.
ELEMENTS OF PLANE GEOMETRY.
137
THEOREM VIII.
260. In any obtuse-angled triangle, the square on the side opposite the obtuse angle equals the sum of the squares of the other two sides plus twice the product of one of those sides and the projection of the other upon that side.
In the A ABC, let c be the obtuse Z-, and PC the pro- jection oi AC upon BC produced.
To prove that AB' =- B C^ + AC' -\- 2 BC X PC.
PB = BC -\- PC. By Algebra, PS = BC' + PC' -\- 2 BC X PC. Add AP^ to each member.
Then AP' -f PB' = BC' + Xp' -j- PC' -{- 2 B C X PC. But AB' = AT' + PB\ and AG'' = A~P' + PC"; (254)
Substitute A B^ and A C^ for their equals. Then ArB"- = BC' -f ^' -{- 2BCX PC. Q. E. D.
12*
138
ELEMENTS OF PLANE GEOMETRY.
THEOREM IX.
261. In any triangle, if a medial line is drawn to the base:
1.— The sum of the squares of the two sides equals twice the square of half the base plus twice the square of the medial line.
II. — Tlie difference of the squares of the two sides equah twice the product of the base and the projection of the medial line upon the base.
In the A ABC, let AD be the medial line, and DP the projection of AD upon the base BC.
To prove
I.— Tliat AB' + AC' = 2 BD' + 2 A^\ II.— That AB" — AC' = 2BCX DP. If AB > ^ C, Z. a is obtuse, and Z. 6 is acute. Then AB' = BD' -f- AD' + 2 J5X> X DP, (260)
and AC' = DC' ^ A~D' — 2DCXDP. (259)
Add these equations, observing that BD = DC. Then jiB' + AC' = 2 BD' + 2 AJf- Subtract the second equation from the first. Then AB' — AC' = 2BCX DP. Q. E. D.
ELEMENTS OF PLANE GEOMETRY. 139
THEOREM X.
262. The sum of the squares of the sides of ayiy quadrilateral equals the sum of the squares of the diagonals plus four times the square of the line joining the middle points of the diagonals.
In the quadrilateral ABCD, let EF be the line joining the middle points of the diagonals JBD and AC. A
D
To prove that
AB' + BG' 4- CW + DA' = AC" + BD" + 4 ET' Draw DE and BE.
Aff -h BC" = 2 AE" -f- 2 BE\ and 7fD' -f ^A' = 2 AE' + 2 ~DE'' (261)
Add these equations.
Then AB'-^BC'-hCD'+ DA = 4 AE'-i- 2 {BK-^DE') Now, 5:e' + :dI;' =^2BF" -{-2 lEF'; (261)
But 4 1:e' = (2 AEf = A~C\
and 4 BF' = (2 jBjP)' = BD';
substituting, JB'^ 4. ;g^2 ^ 05' + ^^' = ^' +
^^^ 4- 4 EF'\ Q. E. D.
263. Cor. — In any parallelogram, the sum of the squares of the sides equals the sum of the squares of the diagonals.
140 ELEMENTS OF PLANE GEOMETRY.
PBOPOETIOJ^AL LIll^ES.
THEOREM XI.
264. If a number of parallels cutting two straight lineSy intercept equal parts on one of the lines, they also intercept equal parts on the other.
Let the lis AB, CD, EF, GH, intersect MN and OP, intercepting on MN equal parts A C, GE, EG.
M 0
To prove that the lis intercept on OP equal parts BD, DF, FH.
Through the points B,D, F, draw BQ, BE, FS, II to 3IN.
AC=BQ,CE = DR, and EG =- FS. (130)
But AC= CE = EG ; (Hyp.)
BQ =DB = FS.
Now, in the As BQD, DBF, and FSH,
/L a = Z. c = Z. e, and /. b = Z. d = A. f; (62)
/\BQD= A DBF = A FSH, (84)
and BD = DF = FH. Q. E. D.
ELEMENTS OF PLANE GEOMETRY.
141
THEOREM XII.
265. A line parallel to the base of a triangle divides the other two sides proportionally.
In the A ABC, let the line DEhe parallel to AB.
To prove that CA : CD :: CB : CE.
Suppose CA and CD to have a common unit of measure, as AG, and suppose it to be contained in CA eight times, and in CD five time.
Then CA : CD :: 8 : 5.
Draw lis to AB through the points of division on CA.
Then CB is divided into eight equal parts, (264)
(164)
Now, this is true, whatever may be the length of the common unit of measure ; hence it is true if it is infinitely- small, as is the case when the lines are incommensurable. Hence in any case the proposition is true. Q. E. D.
|
and |
CB : CE : |
: 8 : 5. |
|
But |
CA : CD : |
: 8 : 5; |
|
, * . |
CA : CD : |
: CB : CE. |
142 ELEMENTS OF PLANE GEOMETRY.
266. Cor. 1. — By (163), the proportion becomes
CA — CD I CA :: CB — CE : CB, or DA : CA :: EB : CB.
267. Cor. 2. — By (161), the last proportion gives
CA : DA :: CB : EB, which by (163)
gives CA — DA : DA :: CB — EB : EB, or CD : DA :: CE : EB;
CD : CE :: DA : EB. (160)
THEOREM XIII.
268. J. straight line which divides two sides of a triangle proportionally is parallel to the third side. .
In the A ABC, let DE divide the sides AB and AC proportionally.
To prove that DE is 11 to BC.
Suppose DE to be drawn II to BC.
Then AB : AD :: AC : AF. (265)
But AB : AD :: AC : AE; (Hyp.)
AC : AF :: AC : AE; (164)
AF = AE, which cannot be unless DF coincides with DE;
DE 13 W to BC, Q.E.D,
ELEMENTS OF PLANE GEOMETRY.
143
SIMILAEITY OF PO
THEOREM XIV
269. Two triangles mutually equiangular are 'si'
Let the Z_s a, 6, c, of the A ABC be respectively equal to the Z-sd, e, /, of the A DEF.
To prove that As ABC and DEF are sit
Place the A ABC on A DEF so as to make Z_ a coin- cide with its equal Z_ d.
Then A ABC takes the position DGH.
Since A 5 = Z. m =- /. ^ G^IT is II to EF ; (64)
Z>^ : ZX^ :: DF : DH,
or DE : ^5 :: 7)i^ : AC. (265)
Likewise we can prove that
DE : ^^ :: EF : J?C.
the A a are similar. (237) Q.E.D.
270. Cor. — Two triangles are similar if two angles of the one are respectively equal to two angles of the other ; two right- angled triangles are similar if one has an acute angle equal to an acute angle of the other.
144
ELEMENTS OF PLANE GEOMETRY.
THEOREM XV.
271. Two triangles whose homologous sides are proportional are similar.
In the As ABC and DEE let
DE : AB :: DF : AC :: EF : BC.
To prove that As ABC and DEE are similar.
On DE and DF, cut off DG and DH respectively equal to AB and A C, and draw GH.
Since DG == AB, and DIT := ^ C,
|
DE : DG :: i)i^ : DH; |
(Hyp.) |
|
GH is W to £i^: |
(268) |
|
Z- a = Z. b, and Z- c = /- d; |
(62) |
|
As DG^IT and i).Ei^ are similar, |
(269) |
|
and DE : DG :: J^Ji^ : GH; |
(237) |
|
also DE : AB :: ^F : BC. |
(Hyp.) |
|
Of these two proportions take the forms |
|
|
DE EF ^ DE EF DG GH'^'''^ AB~ BC |
|
|
Divide; then DG~ GH |
ELEMENTS OF PLANE GEOMETRY.
145
But DG = AB;
GH=BC; A DGH=-A ABC. (86)
But As DGH and DEF are similar;
As ABC and DEF are similar. Q. E. D.
THEOREM XVI.
272. Two triangles having an angle in each equal and the including sides proportional, are similar.
In the As^^C and DEF, let
L. a= Z-h, and DE : AB :: DF : AC.
D
A
To prove that As ABC and DEF are similar.
Place A ABC ox\ A DEF so that Z. a coincides with Z_ h.
B falls somewhere on DE, as at G, and C falls somewhere on DF, as at H.
Then DE : DG :: DF : DH; (Hyp.)
GH i^W to EF. (265)
A c = 21 cZ, and Z_ e = Z. /; (62)
As DG^IT and DEF are similar. (269)
But A DGH = A ABC;
As ^^ O and D^i^ are similar. . Q. E. D. 13
146 ELEMENTS OF PLANE GEOMETRY.
THEOREM XVII.
273. Two triangles having their sides respectively parallel are similar.
In the As ^^C and DEF, let AB, AC, and BC hQ respectively II to DE, DF, and EF.
To prove that ^s ABC and DEF are similar.
Since the sides are II, the corresponding Z_s are either equal or supplements of each other. (68) and (69)
three suppositions can be made :
(1) Z.a4-Z.c?=2U Z.^> + Z.e = 2U Z-c+Z./=2Ls.
(2) Z.a= Z.d, Z. 6 + Z. e = 2 L?, Z. c + Z_/= 2 Ls.
(3)Z.a=Z.(Z, l.b=A-e; .'.Z.c=Z.f.
Now, the sum of all the Z_8 of two ^s cannot exceed 4 Ls;
(76)
the third supposition is the only one admissible; AbABC and DEF are similar. (269) Q. E. P.
ELEMENTS OF PLANE GEOMETRY.
147
THEOREM XVIII.
274. Tivo triangles having their sides respectively perpen- dicular to each other are similar.
In the As ABC and DEF, let the sides AB, AC, and ^0 be respectively J- to DF, FE, and DE.
4 Ls. But
To prove that As ABC and DEF are similar.
Prolong the sides of the A DEF till they meet the sides of the A ABC
The sum of the Z.s of the quadrilateral AGFH equals
(115) Z.S d and e are Ls. (Hyp.)
Z. a + Z. c = 2 Ls. But Z.b + Z.c = 2Ls, (44)
Z.b = /-a. Likewise we can prove
/L 0 = Z- m, and Z- n = Z. p. .'. the As are similar. (269) Q. E. D.
275. Scholium. — In two triangles whose sides are re- spectively parallel, or perpendicular, the homologous sides are the parallel sides, or the perpendicular sides.
148 ELEMENTS OF PLANE GEOMETRY.
THEOREM XIX.
276. Straight lines drawn from the vertex of a triangle to the hose divide the base and its parallel proportionally.
Let CH and CK be straight lines drawn from the vertex C to the base AB of the A ABC, and let DE he parallel to the base.
To prove that
AH : DF :: HK : FG :: KB : GE.
As CAH, CHK, and CKB are respectively similar to As CDF, CFG, and CGE. (269)
.-. AH : DF :: CH : CF :: HK : FG :: CK : CG
:: KB : GE. (237) Q.E.D.
ELEMENTS OF PLANE GEOMETRY.
149
THEOREM XX.
277. Two polygons are similar if they are composed of the same number of triangles similar each to each and similarly placed.
Let the As ABE, EBD, and DEC, of which the polygon ABCDE is composed, be respectively similar to the As FGM, MGK, and KGH, of which the polygon FGHKM is composed.
A B F G
To prove that polygons ABCDE and FGHKM are similar.
Z_ a = Z^ 7n, and l^ h = /- n. (237)
/^ c =^ /- 0,
and /Ld= Z^p. ^ (237)
Add these two equations.
Then l.c-{-Z-d=^/Lo-]r'^P,
or Z. AED = Z. F3IK.
Likewise we can prove Z_ EDO = Z_ MKH, and Z^ ABC = A. FGH;
. • . the polygons are mutually equiangular.
Also AB : FG :: AE : FM :: EB : MG* :: ED :
3fK, etc.; (237)
. • . the polygons have their homologous sides proportional.
.-. polygons ABCDE and FGHKM are similar. (237)
Q.E.D.
13*
150
ELEMENTS OF PLANE GEOMETRY.
THEOREM XXI.
278. Two similar polygons can he divided into the same number of triangles similar each to each and similarly placed.
Let the polygon ABODE be similar to the polygon FGHKMj and let diagonals be drawn from the vertices B and G,
To prove that the As ABE, EBD, and DBG are re- spectively similar to the As FGM, MGK, and KGH.
(237) (272) (237)
A a = Z. fZ, and ^^ : FG :: AE : FM;
^ ABC i^ similar to A FGM.
Also /- h -\- /- c = /- e -{- /- m.
But /. b = /- e;
Z- c = /- m.
And EB : MG :: AE : FM :: ED : MK; (237)
A EBD is similar to A 3IGK. (272)
Likewise we can prove that A DBG ia similar to A KGH.
Q. E. D.
279. Cor. — Two similar polygons can be divided into the same number of triangles similar each to each and similarly placed, by drawing lines to their vertices from any two homolo- gous points.
ELEMENTS OF PLANE GEOMETRY.
151
THEOREM XXII.
280. The perimeters of two similar polygons are to each other as any two homologous sides.
Let the polygon ABODE be similar to the polygon FGHKM, and let p and P denote their perimeters.
To prove that p : P :: AB : FG.
AB : FG :: BC : GH :: CD : HK, etc.; (237)
'. AB-\- BC-\- DC, etc. : FG -{- GH -\- HK, etc. ::
AB : FG, (168)
or
p : P :: AB : FG.
Q.E.D.
281. Cor. — The perimeters of two similar polygons are to each other as any two homologous lines.
152 ELEMENTS OF PLANE GEOMETRY.
THEOREM XXHI.
282. If a perpendicular is drawn from the vertex of the right angle of a right-angled triangle to the hypotheniise :
I. The tivo triangles formed are similar to the given triangle and similar to each other.
II. The perpendicular is a mean proportional between the segments of the hypothenuse.
III. Each side of the right-angled triangle is a mean pro- portional between the hypothenuse and the adjacent segment.
Let ABC be a RA, right-angled at C, and let CD be the _L to the hypothenuse AB.
I. To prove that the /Sb ACD, ABC, and CBD are similar to each other.
Z_ a is common to the RAs A CD and ABC;
A ^07) and A ^^C are similar. (270)
For a like reason As CBD and ABC are similar, and hence similar to A CD.
II. To prove that AD : CD :: CD : DB.
The As ACD and CBD are similar;
AD : CD :: CD : DB. (237)
ELEMENTS OF PLANE GEOMETRY. 153
III. To prove that AB '.AC :: AC : AD, and that AB \ BC :: BC : BD.
The As ACD and ABC are similar;
AB : AC :: AC : AD, (237)
And the As CBD and ABC are similar;
AB : BC :: BC : BD. (237) Q.E.D.
283. Cor. 1. — The squares of the sides about the right angle are to each other as the adjacent segments of the hypothenuse.
From the last two proportions we have
AC'' = ABX AD,
and BC' = AB X BD. (156)
Divide, and cancel the common factor AB.
AC' AD
Then
or
BC BD
AC' : BC' :: AD : BD.
154 ELEMENTS OF PLANE GEOMETRY.
THEOREM XXIV.
284. Two triangles having an angle in each the same are to each other as the products of the sides including the equal angles.
In the ^sABC and DEC let the angle c be common.
C
To prove that £^ ABC : ^DEC :: CAxCB: CDy^ CE.
Draw the line AE.
A ABC : A AEC :: CB : CE,
and A AEC : A DEC :: CA : CD, (249)
Take the forms
A ABC _ CB A AEC~~ CE'
and
A AEC CA
A DEC ~ CD Multiply, and cancel the common factor.
A ABC _ CA X CB ^^ A DEC ~~ CDX CE'
or A ABC : A DEC :: CA X CB : CD X CE. Q.E.D.
ELEMENTS OF PLANE GEOMETRY. 155
THEOREM XXV.
285. If two chords of a circle intersect, their segments are reciprocally proportional.
Let the chords AB and CD intersect at 0.
To prove that CO : AO :: BO : DO. Draw AD and CB.
Z. c = Z- a, and Z. 6 = Z. c?; (206)
As 005 and ilOZ> are similar, (269)
and * CO : AO :: BO : DO. (237) Q.E.D.
156 ELEMENTS OF PLANE GEOMETRY.
THEOREM XXVI.
286. If two secants are drawn from a point without a circle, the secants and their external segments are reciprocally j)ro- portional.
Let the secants PA and PB be drawn from the point P without the O.
To prove that PA : PB :: PE : PD.
Draw AE and BD.
Z. a= Z- b, (206)
and /^ o = Z- o;
/\hPAE and PBD are similar, (270)
and PA : PB :: PE : PD. (237) Q.E.D.
ELEMENTS OF PLANE GEOMETRY. 157
THEOREM XXVII.
287. If a ta7igent and a secant are drawn from a point without a circle^ the tangent is a mean proportional between the secant and its external segment.
Let the tangent FB and the secant PA be drawn from a point P without the O-
To prove that PA : PB :: PB : PC. Draw AB and BC.
Z_ a is measured by 2 arc CB, (206)
and Z. 6 is measured by ^ arc CB; (211)
jL a= /:. b. Also Z. G = Z- G.
As PAB and PBC are similar, (270)
and PA : PB :: PB : PC. (237) Q. E.D.
158
ELEMENTS OF PLANE GEOMETRY.
THEOREM XXVIII.
288. In any triangle, the product of two sides equals the product of the diameter of the circumscribed circle and the perpendicular drawn to the third side from the vertex of the opposite angle.
Let a O be circumscribed about the A ABCy and let CD be a diameter and CE a _L to AB.
To prove that AC X CB = CDX CE.
Draw DB.
l.a= JLd; (206)
RAs AEC and CBB are similar, (270)
and AC \ CD w CE \ CB; (237)
AC XCB = CD X CE. (156) Q.E.D.
ELEMENTS OF PLANE GEOMETRY. 159
THEOREM XXIX.
289. In any triangle the product of two sides equals the product of the segments of the third side formed by the bisector of the opposite angle plus the square of the bisector.
Let a O be circumscribed about the A ABC, and let the Z_ opposite AB be bisected by CD.
C
E
To prove that CB X CA = BD X DA -\- CD\
Produce CD to E, a point in the circumference, and draw AE.
l.b= /- e, (206)
and Z_ c= Z. d; (Cons.)
As GBD and CAE are similar, (270)
and CB '. CE :: CD : CA; (237)
CBX CA = CEXCD. (156)
But CEXCD = {DE-\- CD) CD=:DExCD-{- CD';
CB X CA = DE X CD -\- CD\
Now, DEXCD = BDX DA. (285)
Substitute BD X DA for its equal DE X CD.
Then CB X CA = DB X DA ^ CD\ Q. E. D.
160
ELEMENTS OF PLANE GEOMETRY.
EELATIOISr OF POLTGOISrS.
THEOREM XXX.
290. Similar triangles are to each other as the squares of their homologous sides.
Let the As ABC and EFG be similar.
To prove that AABC: A EFG :: AS" : EF\
Draw the altitudes AD and EH.
Then A ABC : A EFG :: BC X AD :: FG X EH.
(248)
But BC : FG^ :: AB : EF,
and AD : ^il :: ^5 : ^F; (237)
BCxADiFGxFH: AB' : EF' (169)
Compare this with the first proportion; then A^^C: A EFG :: TB' : 'EF\ (164) Q.E.D.
ELEMENTS OF PLANE GEOMETRY.
161
THEOREM XXXI.
291. Similar polygons are to each other as the squares of their homologous sides.
Let the polygons ABODE and FGHKM be similar, and denote their surfaces by s and S.
To prove that s : S :: AB' : FG\
Draw the diagonals A C, EC, and FH, MIT, dividing the polygons into homologous As. (278)
A ABC: AFGH :: AC'
Then A ABC : A FGH :: AC : FH',
and A ^ C^ : A FHM : : A& : FH) (290)
AABC: AFGH :: AACE: AFHM. (164)
Likewise we get A ^ C^ : A FII3f : : A ECD : A MHK;
.-. A ABC ^ A ACE -]- A ECD : AFGH+ A FHM
+ A MHK :: A ABC : A FGH (168)
or s : S :: A ABC : A FGH
But A ABC : A FGH :: AB' : FG'; (290)
s : S :: AB' : FG\ (164) Q.E.D.
292. Cor. — Similar polygons are to each other as the squares of any of their homologous lines.
14*
162
ELEMENTS OF PLANE GEOMETRY.
THEOREM XXXII.
293. If similar polygons are constructed on the three sides of a right-angled triangle, the polygon on the hypothenuse is equivalent to the sum of the polygons on the other two sides.
Let 0, P, and Q be the similar polygons constructed respectively on AB, BC, and AC, the three sides of a RA.
To prove that Q = 0 -\- P.
0 : P :: IB' : BC\ and Q : P ;: AC' : BC\
From the first proportion we get
0 + P : P :: IB' -j- BC" : BC" But AB' + BC' = AC';
0 -\- P : P :: AC' : BC\
(291) (162)
ELEMENTS OF PLANE GEOMETRY. 163
Comparing this with the second proportion, we get Q '. P :: 0 -j- F : P.
Take the form ^ = — p — , and multiply each member by P. Then Q = 0 + P. Q.E.D.
164 ELEMENTS OF PLANE GEOMETRY.
PKOBLEMS IIT CONSTRUCTION
PROBLEM I.
294. To cut a given straight line into any 7iumber of equal parts.
Let AB be the given straight line.
^-=-::n r- ] r 1 iB
c--~-^J I ! I
■-J
i !
->.o
From one extremity A draw the indefinite^ straight line A 0.
Take any convenient line, as A C, and cut it off on A 0 as many times as ^^ is to be cut into equal parts.
Join the last point of division, as P, and the extremity B of the given line.
Through all the other points of division on AO, draw lis to PB.
Then AB is cut into equal parts. (264) Q. E. F.
ELEMENTS OF PLANE GEOMETRY. 165
PROBLEM 11.
295. To cut a given straight line into parts proportional to given straight lines.
Let ABj 0, p, and q be the given straight lines.
E-^O
From A draw the indefinite straight line A 0.
Cut off -40 = 0, CD=p, and DE = q.
Join EB, and draw DG and CF II to EB.
Then the parts AF, FG, and GB are proportional to 0,^, and^. (265) ' Q.E.R
166 ELEMENTS OF PLANE GEOMETRY.
PROBLEM III.
296. To construct a fourth proportional to three given straight lines.
Let n, 0, and p be the three given lines.
|
U |
~ |
F |
||
|
n 0 |
i:-^ |
B |
||
|
_£. |
^^-a |
Construct any convenient Z. , producing the sides AB and A G indefinitely.
Cut off'^D = o,AE = n, and EG == p.
Join ED, and draw GF II to ED.
Then AE : AD :: EG : DF, (267)
or n : 0 :: p \ DF;
DF is the required line. Q. E. F.
297. Scholium.— If EG^ is equal to AD, DF \s the third proportional to AE and AD, or to n and o.
ELEMENTS OF PLANE GEOMETRY.
167
PROBLEM IV.
298. Given two straight lines, to construct a mean proportional between them.
Let 71 and o be the two given lines.
or
Draw the indefinite straight line AE.
Cut off AD = n, and DB = o.
Describe a semi-circumference on AB, and at D erect the -DC.
Draw .4 C and BG.
Z. AGB := a L; (208)
AD : DC :: DC : DB, (282)
n : DC :: DC : o;
DC 13 the required line. Q. E.F.
DEFINITION.
299. A straight line is said to be divided in extreme and mean ratio, when the greater part is a mean proportional between the whole line and the less part.
168
ELEMENTS OF PLANE GEOMETRY.
PROBLEM V.
300. To divide a given straight line in extreme and mean ratio.
Let AB be the given straight line.
/
i)\.'''
O,-'-
:\E
At the extremity B erect the AB.
BC equal to half of
With C as a centre and CB as a radius, describe a O .
Draw J. C, and produce it till it meets the circumference atE.
On AB cut off AP = AD.
Now, AB, being _L to the radius CB at B, is a tangent ;
AE : AB :: AB : AD; (287)
AE — AB : AB :: AB — AD : AD. (163)
But DE = 2 CB = AB;
AE — AB = AD = AP.
AB — AD = AB — AP = PB.
Substitute AP and PB for their equals.
Then AP : AB :: PB : ^P, which gives
AB : AP :: AP : PB; (161)
. AB is divided in extreme and mean ratio at P. (299)
Q.E.F.
ELEMENTS OF PLANE GEOMETRY. 169
PROBLEM VI.
301. On a given straight line to construct a polygon similar to a given polygon.
Let MN be the given line, and ABCDE the given polygon.
Q<
\0
M N
Divide the polygon into As.
At M construct Z. m -- Z_ a, and at N construct /- n = Z. 6.
Then As MNO and ABC are similar. (270)
Likewise construct As MOP and MPQ similar to As ACD and ADE.
Then polygons MNOPQ and ABCDE are similar; (277)
MNOPQ is the polygon required. Q. E. F.
15
170 ELEMENTS OF PLANE GEOMETRY.
PROBLEM VII.
302. To construct a triangle equivalent to a given polygon. Let ABODE be the given polygon.
G
|
''A |
/ |
/ |
||
|
^ / |
\ |
/' |
\ \ |
|
|
// |
\/ |
\ |
// |
— -1 F |
E
Draw the diagonal CAy produce EA, draw BG W to CAj and draw CG.
Then As BAG and GAG have a common base AG and a common altitude;
.-. . ^BAG = ^GAG; (250)
polygon GGDE = polygon ABGDE.
Draw the diagonal GE, produce AE, draw DF II to CEy and draw GF,
Then A FGE = ADGE; (250)
.-. ^ GGF = polygon GGDE = polygon ABGDE;
GGF is the required A. Q. E. F,
ELEMENTS OF PLANE GEOMETRY.
171
PROBLEM VIII.
303. To construct a square equivalent to a given triangle. Let ABC be the given A, 6 its base, and a its altitude.
Find the mean proportional x between a and one-half of h, by (298).
On X construct the square S. Then a:' = ^^ X a = A ABQ;
S is the required square.
(247) Q. E. F.
304. Scholium. — By means of this problem and the pre- ceding one, a square can be constructed equivalent to any given polygon.
172
ELEMENTS OF PLANE GEOMETRY.
PROBLEM IX.
305. To construct a square equivalent to any number of given squares.
Let m, n, o, p, and q be the sides of the given squares.
F
|
/ \ \ |
s m |
1 1 |
|
Of. \ \ |
n |
|
|
1 "^N \ \ |
0 |
|
|
! ""v \ \ A 1 :::^ |
P 5 |
1 1 |
Draw AB = m. At A draw AC = n and At C draw CD == o and At D draw DE = p and At E draw EF = q and
to AB, and draw BC to BC, and draw BD. to jBZ), and draw BE. to ^5, and draw BF.
On JBi^ construct the square S. Now, BF' = EF' -{- BE''
= EF' -^ DE" + BD"
= EF' -^ DE' -{- CD' -i- CB'
= EF'-\- DE' + ~CD' -\- AXf + AF, (254) or S= m^ + w^ + o' 4- j9^ + 5^•
/S is the required square. Q. E. F.
306. Scholium 1. — By means of this and the two pre- ceding problems, a square can be constructed equivalent to the sum of any number of given polygons.
307. Scholium 2. — If m, n, o, p, and q are homologous sides of given similar polygons, BF is the homologous side of a similar polygon which is equivalent to the sum of the given polygons (293).
ELEMENTS OF PLANE GEOMETRY.
173
PROBLEM X.
308. To construct a square equivalent to the difference of two given squares.
Let 0 be the side of the larger square, and p the side of the smaller.
C
M
Construct a L a, and cut AB = p.
With 5 as a centre and o as a radius, describe an arc cutting AM at C.
On A C construct the square S.
Now,
BC" — AB' = AC'
S,
or
S is the required square.
(255) Q. E. F.
309. Scholium 1. — By means of this problem and the two immediately preceding (305), a square can be constructed equivalent to the difference of any two polygons.
310. Scholium 2. — If o and p are the homologous sides of two given similar polygons, AC is the homologous side of a similar polygon equivalent to the difference of the two given polygons.
15*
174
ELEMENTS OF PLANE GEOMETRY.
PROBLEM XI.
311. To construct a rectangle, given its area and the sum of the base and altitude.
Let AB be equal to the given sum of the base and altitude, and let the given area equal that of the square whose side is a.
Ell—
■4-
F \
'B
On AB as a diameter, describe a semicircle, and at A erect
the_L^C = a.
Draw CD li to AB, cutting the circumference at D, and from D draw DP _L to AB.
Now,
DP' = APXPB
(282)
. • . AP is the base and PB is the altitude of the required rectangle. And APFE, whose altitude PF = PB, is the required rectangle. Q. E. F.
ELEMENTS OF PLANE GEOMETRY.
175
PROBLEM XII.
312. To construct a rectangle, given its area and the differ- ence of the base and altitude.
Let AB equal the given difference of the base and altitude, and let the given area be that of the square whose side is a.
C
\
A^
-\B
Yd
On AB as a diameter, describe a O.
At A draw the tangent AC = a, and draw the secant CD so as to pass through the centre 0.
Now, and
CD X CP= CA' = a\ CD —CP=^PD = AB:
(287)
. • . CD is the base and CP is the altitude of the rectangle required, and the rectangle can readily be constructed.
Q. E. F.
176
ELEMENTS OF PLANE GEOMETRY.
PROBLEM XIII.
313. To construct a square having a given ratio to a given square.
Let 0 : J) ho, the given ratio, and S the given square whose side is a.
|
V |
- |
A |
I) / y" 1 / y 1 1 / ' |
||
|
s a |
\ |
||||
|
1 |
F^ |
||||
|
B |
C |
M
On the indefinite line AM, cut off J.^ = o, and BC = p.
On J. C as a diameter, describe a semicircle.
At B erect a _L cutting the circumference at D, and draw AD and CD.
Take DF = a, and draw EF II to AC.
.Now, D^ : DF :: DA : i)C (267), which gives
DE' : DT :: DA' : DC\ (170)
AB : BC :: 0 : p; (283)
But DA' : DC'
DF'
:: 0 : p,
0 : p;
(164)
DE
or DE' : a"
. • . the square whose side is DE is the required one. Q. E.F.
314. Scholium. — Since similar polygons are to each other as the squares of their homologous sides, we can find, by means of the above problem, the homologous side of a poly- gon similar and having a given ratio to a given polygon.
ELEMENTS OF PLANE GEOMETRY. Ill
EXEECISES 1^ H^YENTIOK
THEOREMS.
1. The square inscribed in a circle is equivalent to half the square described on the diameter.
2. Prove geometrically that the square described on the sura of two lines is equivalent to the sum of the squares described on the two lines plus twice the rectangle of the lines.
3. Prove geometrically that the square described on the difference of two lines is equivalent to the sum of the squares described on the two lines minus twice the rectangle of the lines.
4. Prove geometrically that the rectangle of the sum and difference of two lines is equivalent to the difference of the squares described on the lines.
5. If a straight line is drawn from the vertex of an isosceles triangle to any point in the base, the square of this line is equivalent to the rectangle of the segments of the base together with the square of either of the equal sides.
6. The area of a circumscribed polygon equals half the product of the perimeter and the radius of the inscribed circle.
7. The triangle formed by drawing straight lines from the extremities of one of the non-parallel sides of a trapezoid to the middle point of the other, is equivalent to half the trapezoid.
8. The two triangles formed by drawing straight lines from any point within a parallelogram to the extremities of either pair of opposite sides, are equivalent to half the parallelogram.
178 ELEMENTS OF PLANE GEOMETRY.
9. The bisector of the vertical angle of a triangle divides the base into parts proportional to the adjacent sides of the triangle.
PROBLEMS.
1. Trisect a given straight line.
2. Bisect a parallelogram by a line passing through any given point in the perimeter.
3. Construct a parallelogram whose surface and perimeter are respectively equal to the surface and perimeter of a given triangle.
4. On a given straight line construct a rectangle equivalent to a given rectangle.
5. Construct a polygon similar to a given polygon and whose area is in a given ratio to that of the given polygon.
ELEMENTS OF PLANE GEOMETRY.
179
BOOK V.
REGULAE POLYGOI^S AND THE CIRCLE.
DEFINITION.
315. A Regular Polygon is one which is both equilateral and equiangular.
THEOREM h
316. Any equilateral polygon inscribed in a circle is regular. Let F be an equilateral polygon inscribed in a O •
To prove that P is a regular polygon.
Arc AB =: arc BC = arc CD = arc DE, etc.; (189)
arc ABC = arc BCD, etc.; (Ax. 2.)
^b = Z-c = Z.d, etc.; (207)
P is a regular polygon. (315) Q. E. D.
180 ELEMENTS OF PLANE GEOMETRY.
THEOREM n.
317. A circle can be circumscribed about any regular polygon.
Let ABCDEF be a regular polygon.
|
F ^ |
--.7? |
|
|
^\ |
f |
•^ V\ |
|
\ |
0 ") |
|
|
,B^ |
^_--'-^' |
D
To prove that a O can be circumscribed about ABCDEF.
Describe a circumference through the vertices A^ F, and E.
(193)
From 0, the centre, draw OP _L to FE, bisecting it at P, and draw OA and OD.
On OP as an axis, revolve the quadrilateral AOPF till it falls in the plane of ODEP.
Since OP is _L to FE, PF falls on its equal PE, F billing on E; and since /-. a = Z_ b, FA falls on its equal ED^ A falling on D;
OA = OD, and the circumference passes through D.
Likewise we can prove that the circumference passing through F, E, and D passes also through the vertex C, and thus through all the successive vertices of the polygon. Q. E. D.
318. Cor. — A circle can be inscribed in a regular polygon.
ELEMENTS OF PLANE GEOMETRY. 181
DEFINITIONS.
319. The Centre of a regular polygon is the common centre of its circumscribed and inscribed circles.
320. The Angle at tUe Centre of a regular polygon is the angle formed by two lines drawn from the centre to the extremities of any side.
The angles at the centre are equal, any one being equal to four right angles divided by the number of sides of the polygon.
321. The Apotheni of a regular polygon is the perpen- dicular distance from the centre to any side, and is equal to the radius of the inscribed circle.
16
182 ELEMENTS OF PLANE GEOMETRY.
THEOREM III.
322. Regular polygons of the same number of sides are similar figures.
Let p and P be regular polygons of the same number of sides.
E D 0 N
A B G H
To prove that p and P are similar figures.
Since p and P are regular and have the same number of sides, they are mutually equiangular. (114)
Also AB '. BC :: I \ 1,
and GH : HM :: 1 : 1; (315)
AB : BC :: GR : HM; (164)
p and P are similar. (237) Q. E. D.
323. Cor. 1. — The perimeters of similar regular polygons are to each other as the radii of their circumscribed or inscribed circles (281).
324. Cor. 2. — Tlie areas of similar regular polygons are to each other as the squares of the radii of their circumscribed or inscribed circles (292).
ELEMENTS OF PLANE GEOMETRY.
183
EELATION BETWEEIN^ THE CIBCUM-
FEREIsrCE AKD DIAMETER
OF A CIRCLE.
THEOREM IV.
325. The circumferences of circles are to each other as their radii.
Let c and C be the circumferences, r and B the radii, and d and D the diameters of the Os o and 0.
To prove that c : C :: r : B.
Inscribe in the two circles similar regular polygons, and denote their perimeters by p and P.
Then
P
P :: r : R.
(323)
Now, this is true whatever may be the number of sides of the polygons, if there is the same number in each ; hence it is true when the number of sides is infinitely great, in which case p = c, and P =z C, while r and B remain the same;
B.
Q. E. D.
184 ELEMENTS OF PLANE GEOMETRY.
326. Cor. 1. — The circumferences of circles are to each other as their diameters.
By (166) the above proportion becomes
c : C :: 2r : 2B,
or c : C :: d : D.
327. Cor. 2. — The ratio of the circumference of a circle to its diameter is a constant quantity.
By (160) the last proportion becomes
c : d :: C : D,
or c C
H ~d'
This constant ratio is usually denoted by ;r, the Greek letter p, called pi. The numerical value of t: can be found only approximately, as can be proved by the higher mathe- matics.
Hence, in any circle, the circumference and its diameter are incommensurable.
328. Cor. 3. — The ciTcumference of a circle equals the diameter multiplied by r.
C yr = rr, whence C ^= n D,
or C=27z B.
ELEMENTS OF PLANE GEOMETRY.
185
THEOREM V.
329. The area of a regular polygon equals half the product of its perimeter and apothem.
Let P be the perimeter and A the apothem of the regular polygon MNORQS.
To prove that the area of MNORQS = i P X ^.
Draw CO, CR, CQ, etc., dividing the polygon into as many As as it has sides.
All the As have the common altitude A, and the sum of their bases equals P;
the sum of the areas of the As = * P X ^, (247)
or the area of the polygon = i P X ^.
Q. E. D.
16*
186 ELEMENTS OF PLANE GEOMETRY.
THEOREM VI.
330. The area of a circle equals half the product of its circumference and radius.
Let C be the circumference and R the radius of the O 0.
To prove that the area of the circle = ^ C X i?.
Inscribe a regular polygon, and denote its perimeter by P, and its appthem by A.
Then the area of the polygon = i F X A. (329)
Now, this is true whatever may be the number of sides of the polygon; hence it is true when the number is infinitely great, in which case P = C, and A = R;
the area of the O = i C X R- Q.E.D.
331. Cor. 1. — The area of a O ^= t: R\ R heiyig the radius.
The area of a O = ^ C X i^.
But O = 2 t: R; (327)
the area of a O = ?? R^.
332. Cor. 2. — The area of a sector of a circle equals half the product of its arc and the radius.
DEFINITION".
333. Similar Sectors are sectors of different circles, which have equal angles at the centre.
ELEMENTS OF PLANE GEOMETRY. 187
THEOREM VII.
334. Circles are to each other as the squares of their radii. Let r and E denote the radii of the Os o and 0.
To prove that o : 0 :: r^ : R"^
and 0 = n R\ (331)
Divide; then
0 TT r' r'
or 0 : 0 :: r'' : R\ Q.E.D.
335. Cor. — Similar sectors are to each other as the squares of their radii.
188
ELEMENTS OF PLANE GEOMETRY,
PROBLEMS IN COI^^STEUCTIOK
PROBLEM I.
336. To inscribe a square in a given circle. Let 0 be the centre of the given O-
C
Draw any two diameters, as AB and CD, _L to each other, and draw A C, CB, BDj and AD.
Now the angles about the centre are equal ;
the circumference is divided into four equal arcs;
the chords A 0, CB, BD, and AD are equal. (188)
The Z.S ADB, DBC, BCA, and CAD are Ls; (208)
ADBC is the required square. (123) Q. E. F.
337. Cor. — To inscribe a regular polygon of 8 sides, bisect the arcs subtended by the sides of an inscribed square and draw chords; and by continuing the process, we can inscribe regular polygons of 16, 82, etc., sides.
ELEMENTS OF PLANE GEOMETRY. 189
PROBLEM 11.
338. To inscribe a regular hexagon in a circle. Let 0 be the centre of a given O.
Draw any radius, as OA, and with J. as a centre and the radius of the circle describe an arc, cutting the circumference at B.
Draw AB and OB.
Now, the A AB 0 is both equilateral and equiangular ; (97)
Z. a = i of 2 Ls = i of 4 Ls; (81)
. • . arc AB = i of the circumference, and the chord AB is the side of a regular inscribed hexagon ;
. • . ABCDEF, which is formed by applying the radius six times as a chord, is the required hexagon. Q. E. F.
339. Cor. 1. — To inscribe an equilateral triangle, join the alternate vertices of a regular inscribed hexagon.
340. Cor. 2. — To inscribe a regular polygon of 12 sides, bisect the arcs subtended by the sides of a regular inscribed hexagon and draw chords; and by continuing the process, we can inscribe regular polygons of 2J^, Jf8, etc., sides.
190 ELEMENTS OF PLANE GEOMETRY,
PROBLEM III. 341. In a given circle to inscribe a regular decagon. Let 0 be the centre of the given O .
Suppose the problem to be solved, and let ABC^ etc., be the regular inscribed decagon.
Draw ^C and ^i).
Now, AC and BD bisect the circumference;
they are diameters and intersect at the centre C.
Draw BEy cutting ^ (7 at P.
Z_ a is measured by h (arc AB + arc EC), or g arc BC,
(212) and Z^ b is measured by ^ arc BC; (206)
A APB is isosceles, and AB = BP,
ELEMENTS OF PLANE GEOMETRY. 191
Also, Z_ c^ is measured by i arc ED, or arc AB, and Z_ e is measured by arc AB; , (205)
A .BOP is isosceles, and OP = BP = AB. Z_ c is measured by ^ arc AE, or AB; (206)
. * . As APB and ABO are mutually equiangular and similar;
(269) AO : AB :: AB : AP,
or AO : OP :: OP : AP.
But this shows that A 0, the radius, is divided in extreme and mean ratio at P, and that OP, the greater part, equals AB, Si side of the regular inscribed decagon.
Therefore, to inscribe a regular decagon, divide the radius in extreme and mean ratio, and apply the greater part ten times as a chord. Q. E. F.
342. Cor. 1. — To inscribe a regular pentagon, join the alternate vertices of a regular inscribed decagon.
343. Cor. — To inscribe a regular polygon of 20 sides, bisect the arcs subtended by the sides of a regular inscribed decagon and draw chords; and by continuing the process, we can inscribe regular polygons of 40, 80, etc., sides.
192 ELEMENTS OF PLANE GEOMETRY.
PROBLEM IV.
344. In a given circle to inserihe a regular pentadecagon. Let C be the given O-
Draw the chord AB equal to the side of a regular inscribed hexagon, and the chord BD equal to the side of a regular inscribed decagon, and draw A D.
Now, arc AD = arc AB — arc DB
= i of the circumference — J^ of the circumference
= y^^ of the circumference.
Therefore chord AD = a side of a regular inscribed penta- decagon; and hence if we apply AD fifteen times as a chord we get the required polygon. Q. E. F.
345. Cor. — To inscribe a regular polygon of SO sides, bisect the arcs subtended by the sides of a regular inscribed pentadecagon and draw chords; and by continuing the process, we can inscribe regular polygons of 60, 120, etc., sides.
ELEMENTS OF PLANE GEOMETRY. 193
PROBLEM V.
346. In a circle whose radius is unity, to find the value of the chord of half an arc in terms of the chord of the whole arc.
Let 0 be the centre of a O whose radius is 1, AB the chord of an arc, and BC the chord of half the arc.
0
Draw the radii OB and OC.
In the RA BDO, OB' = OD' + BB^.
Whence OD = Vd^ — BI)\
But OB --=00 = 1, and BD = ^;
In the UaCDB,BC= V^Wff~-\- CD\
But CD = 1 — 02) - 1 -Vl — |- |-
Substitute -^ and 1 — V 1 — Hr ior then- equals
2 '^"^^ ^ ' ^ \ 2 / BD and CD, and reduce
Then BC = y2 — 1/4 — ^^l Q. E. F.
17
194 ELEMENTS OF PLANE GEOMETRY.
PROBLEM VI.
347. To find the numerical value of iz, approximately. Let C be the circumference, and R the radius of a O.
When B = 1
(327)
Now, by means of the formula BC = \/2 — VT--A^\ established in (346), we make the following computations :
In a regular inscribed polygon of No. Sides. Form of Computation. Length of Side. Perimeter.
6. See (338) 1.00000000 6.00000000.
12. jB 0=1/2—1/4 — r = .51763809 6.21165708.
24. BC= \/2--/4— (.517638^^== .26105238 6.26525722.
48. 5C=l/2— 1/4^261052387^= .13080626 6.27870041.
96. 50=1/2—1/4— (.13080626)'^= .06543817 6.28206396.
192. 50= V^2— 1/4— (.065"43817)^= .03272346 6.28290510.
384. 50= 1^2— 1/4— (.03272346^= .01636228 6.28311544.
768. 50= l/2-l/4^=(jai6362l8y2= .00818121 6.28316941.
It will be seen that the first four decimal places remain the same, to whatever extent we increase the number of sides. Hence we can consider 6.28317 as the approximate value of the circumference of a circle whose radius is 1 .
^ ^ O 6^2^ ^ g^^^g ^^^^,jy Q ^ -p
ELEMENTS OF PLANE GEOMETRY. 195
EXEECISES IIsT il^YEl^TIOl^.
THEOREMS.
1. The side of an inscribed equilateral triangle equals half the side of the circumscribed equilateral triangle.
2. The diameter of a circle is a mean proportional between the sides of the equilateral triangle and the regular hexagon circumscribed about the circle.
3. The square inscribed in a circle equals half the square on the diameter.
4. The area of a regular inscribed hexagon equals three- fourths the area of a regular circumscribed hexagon.
5. The area of a regular inscribed hexagon is a mean proportional between the areas of the inscribed and circum- scribed equilateral triangles.
6. If the vertices of a square are taken as centres and half the diagonal as a radius and circles be described, the points of intersection of the circumferences and the sides of the square are the vertices of a regular octagon.
7. The area of a regular inscribed octagon equals the area of a rectangle whose adjacent sides equal the sides of the circumscribed and inscribed squares.
8. The area of a regular inscribed dodecagon equals three times the square on the radius.
196 ELEMENTS OF PLANE GEOMETRY,
PROBLEMS.
1. Inscribe in a given circle a regular polygon similar to a given regular polygon.
2. Circumscribe a polygon similar to a given inscribed polygon. ,
3. In a given circle, inscribe three equal circles, touching each other and the given circle.
4. In a given circle, inscribe four equal circles in mutual contact with each other and the given circle.
5. In a given equilateral triangle, inscribe three equal circles, touching each other, and each touching two sides of the triangle.
6. About a given circle, describe six circles, each equal to the given one and in mutual contact with each other and the given circle.
O
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