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on the same side, GHD, and the two interior angles on the same side, BGH, GHD, shall be together equal to two right angles.

For if the angle AGH be not equal to the angle GHD, one of them must be greater than the other; let the angle AGH be the greater.

Then the angle AGH is greater than the angle GHD;

to each of them add the angle BGH;

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therefore the angles AGH, BGH are greater than the angles BGH, GHD.

But the angles AGH, BGH are together equal to two right angles;

[I. 13. therefore the angles BGH, GHD are together less than two right angles.

But if a straight line meet two straight lines, so as to make the two interior angles on the same side of it, taken together, less than two right angles, these straight lines being continually produced, shall at length meet on that side on which are the angles which are less than two right angles. [Axiom 12. Therefore the straight lines AB, CD, if continually produced, will meet.

But they never meet, since they are parallel by hypothesis. Therefore the angle AGH is not unequal to the angle GHD; that is, it is equal to it.

But the angle AGH is equal to the angle EGB. [I. 15. Therefore the angle EGB is equal to the angle GHD. [Ax. 1. Add to each of these the angle BGH.

Therefore the angles EGB, BGH are equal to the angles BGH, GHD. [Axiom 2.

But the angles EGB, BGH are together equal to two right angles.

[I. 13.

Therefore the angles BGH, GHD are together equal to two right angles.

Wherefore, if a straight line &c. Q.E.D.

[Axiom 1.

PROPOSITION 30. THEOREM.

Straight lines which are parallel to the same straight line are parallel to each other.

Let AB, CD be each of them parallel to EF: AB shall be parallel to CD.

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angle AGK is equal to the angle GHF.

Therefore the angle AGK is equal to the angle GKD; [Ax. 1.

and they are alternate angles;

therefore AB is parallel to CD.

Wherefore, straight lines &c. Q.E.D.

PROPOSITION 31. PROBLEM.

[I. 27.

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 draw a straight line through the, point A parallel to the straight line BC.

In BC take any point D, and join AD; at the point A in the straight line AD, make the angle DAE equal to the alternate angle ADC; [I. 23.

E

B

and produce the straight line EA to F.

EF shall be parallel to BC.

Because the straight line AD, which meets the two straight lines BC, EF, makes the alternate angles EAD, ADC equal to one another,

EF is parallel to BC.

[Construction.

[I. 27.

Wherefore the straight line EAF is drawn through the given point A, parallel to the given straight line BC. Q.E.F.

PROPOSITION 32. THEOREM.

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 together 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 shall be equal to the two interior and opposite angles CAB, ABC; and the three interior angles of the triangle, namely, ABC, BCA, CAB shall be equal to two right angles.

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Again, because AB is parallel to CE, and BD falls on them, the exterior angle ECD is equal to the interior and opposite angle ABC.

[I. 29. 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. [Axiom 2. To each of these equals add the angle ACB; therefore the angles ACD, ACB are equal to the three angles CBA, BAC, ACB. [Axiom 2. But the angles ACD, ACB are together equal to two right angles; therefore also the angles CBA, BAC, ACB are together equal to two right angles.

Wherefore, if a side of any triangle &c.

[I. 13,

[Axiom 1.

Q.E.D.

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

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 of these triangles are equal to twice as many right angles as there are triangles, that is, as the figure has sides.

And the same angles are equal to the interior angles of the figure, together with the angles at the point F, which is the common vertex of the triangles,

that is, together with four right angles. [I. 15. Corollary 2. Therefore all the interior angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides.

COROLLARY 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 angle ABD, is equal to two

right angles;

[I. 13.

therefore all the interior angles

of the figure, together with all
its exterior angles, are equal to D
twice as many right angles as

the figure has sides.

B

But, by the foregoing Corollary all the interior angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides.

Therefore all the interior angles of the figure, together with all its exterior angles, 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 and CD be equal and parallel straight lines, and let them be joined towards the same parts by the straight lines AC and BD: AC and BD shall be equal and parallel.

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and BC is common to the two triangles ABC, DCB; the two sides AB, BC are equal to the two sides DC, CB, each to each;

and the angle ABC was shewn to be equal to the angle BCD ;

therefore the base AC is equal to the base BD, and the triangle 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.

[I. 4.

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.

Wherefore, the straight lines &c. Q.E.D.

PROPOSITION 34. THEOREM.

[I. 27.

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

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

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