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For, if BC be not equal to EF, let BC be the greater of them, and make BH equal to EF, and join AH; and because BH is equal to EF, and AB to DE ; the two AB, BH are equal to the two DE, EF, each to each ; and they contain equal angles; therefore the base AH is equal to the base DF, and the triangle ABH to the triangle DEF, and the other angles shall be equal, each to each, to which the equal sides are opposite ; therefore the angle BHA is equal to the angle EFD ; but EFD is equal to the angle BCA, therefore also the angle BHA is equal to the angle BCA; that is, the exterior angle BHA of the triangle AHC is equal to its interior and opposite angle BCA; which is impossible ; (16. 1.) wherefore BC is not unequal to EF, that is, it is equal to it; and AB is equal to DE; therefore, the two AB, BC are equal to the two DE, EF, each to each; and they contain equal angles; wherefore the base AC is equal to the base DF, and the third angle BAC to the third angle EDF. Therefore, if two triangles, &c. Q. E. D.

Proposition XXVII. Theorem. If a straight line falling upon two other straight lines makes the alternate angles equal to one another, these two straight lines shall be parallel

Let the straight line EF, which falls upon the two straight lines AB, CD make the alternate angles AEF, EFD, equal to one 2..other; AB is parallel to CD.

For, if it be not parallel, AB and CD being produced shall meet either towards B, D), or towards A, C : let them be produced and meet towards B, D in the point G; therefore GEF is a triangle, and its exterior angle AEF is greater (16. 1.) than the interior and opposite angle EFG; but it is also equal to it, which is impossible ; therefore AB and CD being produced do not meet towards B, D.

In like manner it may be demonstrated, that they do not meet towards A, C; but those straight lines which meet neither way, though produced ever so far, are parallel (35 Def.) to one another. AB therefore is parallel to CD. Wherefore, if a straight line, &c. Q. E. D.

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Proposition XXVIII. Theorem. If a straight line falling upon two other straight lines makes the exterior angle equal to the interior and opposite upon the same side of the line ; or makes the interior angles upon the same side together equal to two right angles; the two straight lines shall be parallel to one another.

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Let the straight line EF, which falls upon the two straight lines AB, CD, make the exterior angle EGB equal to the interior and opposite angle GHD upon the same side; or make the interior angles on the same side BGH, GHD together equal to two right angles ; AB is parallel to CD.

Because the angle EGB is equal to the angle GHD, and the angle EGB equal (15. 1.) to the angle AGH, the angle AGH is equal to the angle GHD; and they are the alternate angles ; therefore AB is parallel (27. 1.) to CD.

Again, because the angles BGH, GHD are equal (By Hip.) to two right angles; and that AGH, BGH are also equal (13. i.) to two right angles; the angles AGH, BGH are equal to the two angles BGH, GHD.

Take away the common angle BGH; therefore the remaining angle AGH is equal to the remaining angle GHD; and they are alternate angles ; therefore AB is parallel to CD. Wherefore if a straight line, &c. Q. E. D.

H

Proposition XXIX. Theorem. If a straight line fall upon two parallel straight lines, it makes the alternate angles equal to one another ; and the exterior angle equal to the interior and opposite upon the same side ; and likewise the two interior angles upon the same side together equal to two right angles.

Let the straight line EF fall upon the parallel straight lines AB, CD; the alternate angles AGH, GHD are equal to one another ; and the exterior angle EGB is equal to the interior and opposite, upon the same side, GHD; and the two interior angles BGH, GHD upon

the same side, are together equal to two right angles.

B For, it AGH be not equal to GHD, one of them must be greater than the

D other ; let AGH be the greater ; and because the angle AGH is greater than the angle GHD, add to each of them the angle BGH; therefore the angles AGH, BGH are greater than the angles BGH, GHD; but the angles AGH, BGH are equal (13. 1.) to two right angles ; therefore the angles BGH, GHD; are less than two right angles ; but those straight lines which, with another straight line falling upon them, make the interior angles on the same side less than two right angles, do meet (12. Ar.) together if continually produced ; therefore the straight lines AB, CD, if produced far enough shall meet; but they never meet, since they are parallel by the 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 (15. 1.) to the angle EGB; therefore likewise EGB is equal to GHD; add to each of these the angle

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BGH ; therefore the angles EGB, BGH are equal to the angles BGH, GHD; but EGB, BGH are equal (13. 1.) to two right angles; there. fore also BGH, GHD are equal to two right angles. Wherefore, if a straight, &c. Q. E. D.

Proposition XXX. Theorem.

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, CL, and because GHK cuts the parallel straight lines AB, EF, the angle AGH is equal (29. 1.) to the Aangle GHF. Again, because the straight line GK cuts the parallel straight lines EF, CD, the angle GHF is equal (29. 1.) to the 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 (27. 1.) to CD. Wherefore straight lines, &c. Q. E. D.

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B

D

Proposition XXXI. Problem. 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; and at the point A,' in the straight line AD, make (23. 1.) the angle DAE equal to the angle ADC ; and produce the straight line EA to F.

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 (27. 1.) to BC. Therefore the straight line EAF is drawn through the given point A, parallel to the given straight line BC. Which was to be done.

Proposition XXXII. 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 equal to two right angles.

Let ABC be a triangle, and let one of its sides BC be produced to

2

B

D

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D; the exterior angle ACD is equal to the two interior an 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 pa-

E.
rallel (31. 1.) to the straight line AB ;
and because AB is parallel to CE, and
AC meets them, the alternate angles
BAC, ACE are equal (29. 1.)

Again, because AB is parallel 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 angie BAC; therefore the whole exterior angle ACD is equal to the two interior and opposite angles CAB, ABC; to these equals 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 (13. 1.) 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.

Cor. 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 di. vided 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 there are sides of the figure, and the same angles are equal to the angles of the figure, together with the angles at the point F, which is the common vertex of the triangles : that is, (2. Cor. 15. 1.) together with four right angles. Therefore all the angles of the figure, together with four right angles, are equal to twice as many right ang.es as the figure has sides.

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 equal (13. 1.) to two right angles ; therefore all the interior together with all the exterior angles of the figure, are equal to twice as many right angles as there are sides of the figure ; that is, by the foregoing 0 B corollary, they are equal to all the interior angles figure, together with four right angles ; therefore all the exterior angles are equal to four right angles,

Proposition XXXIII. Theorem. The straight lines wbich join the extremities of two equal and perallel straight lines, towards the same parts, are also themselves equal and parallel.

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Let AB, CD be equal and parallel

B straight lines, and joined towards the same parts by the straight lines AC, BD; AC, BD are also equal and parallel.

D Join BC; and because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD are equal ; (29. 1.) and because AB is equal to CD, and BC common to the two triargles ABC, DCB, the iwo 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 (4. 1.) to the base BD, and the triangle ABC to the triangle BCD, and the other angles to the other angles, (4.1.) 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 (27.1.) to BD,

and it was shewn to be equal to it. Therefore, straight lines, &c. Q.E.D.

Proposition XXXIV. Theorem.

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The opposite sides and angles of parallelograms are equal to one another, and the diameter bisects them, that is, divides them 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 ABCD be a parallelogram, of which BC is a diameter ; the opposite sides and angles of the figure are equal to one another; and the diameter BC bisects it.

Because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD, are equal (29. 1.) to one another; and because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal (29. 1.) to one another ; wherefore the two triangles ABC, CBD have two angles ABC, BCA in one, equal to two angles BCD, CBD in the other, each to each, and one side BC common to the two triangles, which is adjacent to their equal angles ; therefore their other sides shall be equal, each to each, and the third angle of the one to the third angle of the other, (26. 1.) viz. the side (26 1.) AB to the side CD, and AC to BD, od the angle BAC equal to the angle BDC.

And because the angle ABC is equal to the angle BCD, and the angle SBD to the angle ACB, the whole angle ABD is equal to the whole ingle ACD : And the angle BAC has been shewn to be equal to the angle BDC : therefore the opposite sides and angles of parallelograins are equal to one another ; also, their diameter bisects them ; for AB being equal to CD, and BC common, the two AB, BC are equal to the (wo DC, CB, each to each ; and the angle ABC is equal to the augle

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