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angle upon the same side of the straight line and the two interior angles upon the same side of it together equal to two right angles. And conversely, if the exterior angle is equal to the interior apposite angle or the two interior angles are together equal to two right angles; the two straight lines are parallel.

Let EF fall upon AB, CD which are parallel. The exterior angle EGB is equal to the interior and opposite angle GHD upon the same side of the straight line EF. And the two interior angles

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BGH, GHD upon the same side of it are together equal to two right angles.

Because the angle AGH is equal (P. 2) to the anglo EGB, but the angle AGH is equal to the angle GHD (P. 7) therefore the angle EGB is equal (Ax. 1) to the angle GHD.

Again, to each of these equals, add the angle BGH. Therefore the two angles EGB, BGH, are equal (Ax. 2) to the two angles BGH, GHD.. But the two angles EGB, BGH are equal (cor 2: P. 5) to two right angles. Therefore also the two angles BGH, GHD are equal (Ax. 1) to two right angles.

Conversely, let the exterior angle EGB be equal to the interior opposite angle GHD. Because the angle AGH is equal to the angle EGB, therefore is also equal (Ax. 1) to GHD the alternate angle; therefore AB, CD are parallel (P. 7).

Lastly, let the two interior angles BGH and GHD be together equal to two right angles. But BGH and AGH are together equal to two right angles (cor. 2:5). Therefore the angles GHD and AGH (Ax. 3) are equal, Because the alternate angles are equal, therefore AB

and CD are parallel (P.7). Wherefore if a straight line &c. Q. E. D.

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

PROP. IX. PROBLEM. (E. 1. 31.)

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

Let A be a point and GH a straight line. It is required to draw, through A, a straight line parallel to GH.

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H

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B

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Take any two points B and C in GH. Join AB, AC. Bisect AC at F (P. 5). Produce BF to D and make FD equal to BF (Def. 6), Join AD.

Because in the two triangles AFD and BFC, the two sides AF, DF are equal to the two sides CF, BF and the angles AFD and BFC are also equal (P. 2). Therefore the angle FAD and FCB are equal (P. 3).

Because the straight lines CA, falling upon the two other straight lines AD and GH makes the alternate angles CAD and ACG equal, therefore these two straight lines are parallel (Prop. 7). Wherefore, through the point A, AD is drawn parallel to the straight line GII. Q. E. F.

Cor. Hence a perpendicular can be drawn to a straight line or to a straight line produced from a point either within or without the straight line.

Because, a perpendicular can be drawn from the point of bisection (P. 5) and a straight line parallel to it can be drawn from the given point (P. 9). The parallel straight line will be the perpendicular required.

PROP. X. THEOREM. (E. 1. 26.)

If two triangles have two angles of the one equal to treo angles of the other, each to each, and one side equal to one side;-viz. either the sides adjacent to the equal angles, or the sides opposite to the equal angles in each; then their other sides are equal, each to each, and also the third angle of the one to the third angle of the other.

Let ABC, DEF be two triangles which have the two angles ABC and BCA of the one equal to the two angles DEF, EFD of the other, each to each; viz., ABC to DEF and BCA to EFD. Also a side of the one triangle equal to a side of the other triangle. First, let those sides be equal, which are adjacent to the angles that are equal in the two triangles; viz., BC to EF. Then their other sides are equal, each to each; viz., AB to DE, and AC to DF; and the third angle BAC is equal to the third angle EDF.

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Place the triangle DEF upon the triangle ABC; so that the base EF may coincide with the base BC. Then the angle DEF will coincide with the angle ABC and the angle DFE with the angle ACB, because they are equal. Because, the angle DEF coincides with the angle ABC, therefore the side DE is placed on the side AB and they wholly coincide (Ax. 12). For the same reason, the side DF wholly coincides with the side AC. But if the angle EDF do not coincide with the

angle BAC, then also the sides DE and DF do not wholly coincide with the sides AB and AC: while it has been shown that they wholly coincide. Therefore the angles EDF and BAC coincide. Wherefore the sides DE, DF and AB, AC are equal and so are the angles EDF and BAC.

Next, let those sides which are opposite to the equal angles in each triangle be equal to one another; viz, AB to DE. Then their other sides are equal; viz., AC to DF, and BC to EF. And the third angle BAC is equal to the third angle EDF.

Apply the triangle DEF to the triangle ABC; so that the side DE may be on the side AB and the angle DEF coincide with the angle ABC. Therefore the side EF is placed on the base BC and they coincide and the triangle DEF coincides with the triangle ABC.

For, if not, let the two triangles ABC and DEF stand as ABC and ABF. But this is impossible. Because the angle AFB (ie., DFE) is greater than the interior opposite angle ACF (P. 6), while they have been shown to be equal (hyp.) Wherefore the two triangles coincide and are therefore equal (Ax. 6). Therefore the two sides AC, BC are equal to the two sides DF, EF, each to each. And the angle EDF is equal to the angle BAC. Wherefore, if two triangles, &c. Q. E. D.

PROP. XI. THEOREM. (E. I. 34)

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

Let AC be a parallelogram, of which BD is a diagonal. The opposite sides and angles of the figure are equal to e another; and the diagonal BD bisects it.

Because AB is parallel to CD, and BD meets them, the angle ABD is equal (P. 7) to the alternate angle BDC.

Because AD is parallel to BC, and BD meets them, the angle ADB is equal (P. 7) to the alternate angle CBD. Because in the two triangles ABD, CBD, the two angles ABD, BDA, in the one,

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are equal to the two angles BDC, CBD in the other, each to each; and one side BD, adjacent to these equal angles, is common to the two triangles. Therefore their other sides are equal, each to each, and the third angle of the one is equal to the third angle of the other (P. 10); viz., the side AB to the side CD, the side AD to the side BC, and the angle BAD to the angle BCD. Because the angle ABD is equal to the angle BDC, and the angle CBD to the angle ADB. Therefore the whole angle ABD is equal (Ax. 2) to the whole angle ADC. And the angle BAD has been proved to be equal to the angle BCD. Therefore the opposite sides and angles of a parallelogram are equal to one another.

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Also the diagonal BD bisects the parallelogram AC, because the two triangles ABD, BCD, are equal. Similarly it can be proved that the diagonal AC bisects the parallelogram. Wherefore the opposite sides &c. Q.E.D.

Cor. Straight lines which join the extremities of two equal parallel straight lines are equal and parallel (E. 1. 33).

PROP. XII. THEOREM. (E. I. 35).

Parallelograms upon the same base, and between the same parallels, are equal to one another.

Let the parallelograms AC, BF be upon the same base

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