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For if AGH be not equal to GHD, let KG be drawn Book I. making the angle KGH equal to GHD, and produce

KG to L; then KL will be parallel to CDa: but AB is also parallel to CD; thereforetwo straight lines K. are drawn through

A

E

L

a 27. 1.

B

the same point G, C

parallel to CD, and

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which is impossible. The angles AGH, GHD there- b 11. Ax. fore are not unequal, that is, they are equal to one another. Now, the angle EGB is equal to AGH; and c 15. 1. AGH is proved to be equal to GHD; therefore EGB is likewise equal to GHD; add to each of these the angle BGH; therefore the angles EGB, BGH are equal to the angles BGH, GHD; but EGB, BGH are equal to two d 13. 1. right angles; therefore also BGH, GHD are equal to two right angles. Wherefore, if a straight line, &c. Q.

E. D.

COR. If two lines KL and CD make, with EF, the two angles KGH, GHC together less than two right angles, KG and CH will meet on the side of EF on which the two angles are that are less than two right angles.

For, if not, KL and CD are either parallel, or they meet on the other side of EF; but they are not parallel; for the angles KGH, GHC would then be equal to two right angles. Neither do they meet on the other side of EF; for the angles LGH, GHD would then be two angles of a triangle, and less than two right angles; but this is impossible; for the four angles KGH, HGL, CHG, GHD are together equal to four right angles d, of which the two, KGH, CHG are by supposition less than two right angles; therefore, the other two, HGL, GHD are greater than two right angles. Therefore, since KL and CD are not parallel, and since they do not meet towards L and D, they must meet if produced towards K and C.

Book I.

PROP. XXX. THEOR.

Straight lines which are parallel to the same straight line are parallel to one another.

Let AB, CD, be each of them parallel to EF; AB is also parallel to CD.

Let the straight line GHK cut AB, EF, CD; and because GHK cuts the

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b 27. 1.

a

to the angle

GHF. Again, because
the straight line GK cuts E
the parallel straight lines

A

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angle GKD; and it was

shewn that the angle

AGK is equal to the angle GHF; therefore also AGK is equal to GKD; and they are alternate angles; therefore AB is parallel to CD. Wherefore straight lines, &c. Q. E. D.

b

PROP. XXXI. PROB.

To draw a straight line through a given point parallel to a given straight line.

Let A be the given point, and BC the given straight line; it is required to

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draw a straight line E

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A

F

C

a 23. 1.

b 27. 1.

In BC take any point

D, and join AD; and at the point A, in the straight line AD, make a the angle DAE equal to the angle ADC; and produce the straight line EA to F.

Because the strai ine AD, which meets the two straight lines BC, Eakes the alternate angles EAD, ADC equal to one another, EF is parallel to BC. Therefore the straight line EAF is drawn through the given

point A parallel to the given straight line BC. Which Book I.. was to be done.

a

PROP. XXXII. THEOR.

If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are equal to two right angles.

Let ABC be a triangle, and let one of its sides BC be produced to D; the exterior angle ACD is equal to the two interior and opposite angles CAB, ABC; and the three interior angles of the triangle, viz. ABC, BCA, CAB, are together equal to two right angles.

Through the point

C draw CE parallel

to the straight line AB; and because AB is parallel to CE and AC meets them,

the alternate angles

B

A

a 31. 1.

E

D

BAC, ACE are equal". Again, because AB is pa- b 29. 1. rallel to CE, and BD falls upon them, the exterior angle ECD is equal to the interior and opposite angle ABC; but the angle ACE was shewn to be equal to the angle BAC; therefore the whole exterior angle ACD is equal to the two interior and opposite angles CAB, ABC; to these angles add the angle ACB, and the angles ACD, ACB are equal to the three angles CBA, BAC, ACB; but the angles ACD, ACB are equal to two right angles; therefore also the angles CBA, BAC, ACB are equal to two right angles. Wherefore, if a side of a triangle, &c. Q. E. D.

c

COR. 1. All the interior angles of any rectilineal figure are equal to twice as many right angles as the figure has sides, wanting four right angles.

For any rectilineal figure ABCDE can be divided into as many triangles as the figure has sides, by drawing straight lines from a point F within the figure to each of its angles. And, by the preceding proposition, all the angles

D

c 13. 1.

Book 1. of these triangles are equal to
twice as many right angles as
there are triangles, that is, as
there are sides of the figure;
and the same angles are equal E
to the angles of the figure, to-
gether with the angles at the
point F, which is the common
vertex of the triangles: that
a 2. cor. is a, together with four right

15, 1.

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angles. Therefore, twice as many right angles as the figure has sides, are equal to all the angles of the figure, together with four right angles, that is, the angles of the figure are equal to twice as many right angles as the figure has sides, wanting four.

COR. 2. All the exterior angles of any rectilineal figure are together equal to four right angles.

Because every interior angle ABC, with its adjacent exterior ABD, is

b 13. 1. equal b to two right

angles; therefore

*

A

all the interior, to-
gether with all the
exterior angles of

the figure, are e-D
qual to twice as
many right angles
as there are sides

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of the figure; that is, by the foregoing corollary, they are equal to all the interior angles of the figure, together with four right angles; therefore all the exterior angles are equal to four right angles.

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The straight lines which join the extremities of two equal and parallel straight lines, towards the same parts, are also themselves equal' and parallel.

Let AB, CD, be equal and parallel straight lines, and joined towards the same parts by the straight lines

AC, BD; AC, BD are al-A

so equal and parallel.

Join BC, and because AB

is parallel to CD, and BC meets them, the alternate angles ABC, BCD are equal a:

and because AB is equal to

b

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CD, and BC common to the two triangles ABC, DCB, the two sides AB, BC are equal to the two DC, CB; and the angle ABC is equal to the angle BCD; therefore the base AC is equal to the base BD, and the tri- b 4. 1. angle ABC to the triangle BCD, and the other angles to the other angles", each to each, to which the equal sides are opposite; therefore the angle ACB is equal to the angle CBD; and because the straight line BC meets the two straight lines AC, BD, and makes the alternate angles ACB, CBD equal to one another, AC is parallel to BD; and it was shewn to be equal to it. c 27. 1. Therefore, straight lines, &c. Q. E. D.

PROP. XXXIV. THEOR.

The opposite sides and angles of a parallelogram are equal to one another, and the diameter bisects it, that is, divides it into two equal parts.

N. B.-A Parallelogram is a four-sided figure, of which the opposite sides are parallel; and the diameter is the straight line joining two of its opposite angles.

Let ACDB be a parallelogram, of which BC is a diameter; the opposite sides and angles of the figures are equal to one another; and the diameter BC bisects it. Because AB is parallel to A

CD, and BC meets them, the alternate angles ABC, BCD are equal a to one another; and because AC is parallel to BD and BC meets them, the alternate

angles ACB, CBD are equal a to one another; wherefore the two triangles ABC, CBD have two angles ABC,

C

a 29. 1.

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