the third angle BAC to the third angle EDF. Therefore, if two triangles, &c. Q.E.D. PROP. XXVII. THEOR. lines makes the alternate angles equal to one Let the straight line EF, which falls upon the two straight lines AB, CD, make the alternate angles AEF, EFD equal to one another: AB shall be parallel to CŨ. For, if it be not parallel, AB and CD being produced will 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 * 16. 1. triangle, and its exterior angle AEF is greater* than the interior and opposite angle EFG ; but it + Hyp. is also equals to it, which is impossible; therefore AB A E B G but those straight lines which meet neither rway, *35 Def, though produced ever so far, are parallel * to one another: therefore AB is parallel to CD. Wherefore, if a straight line, &c. Q. E.D. PROP. XXVIII. THEOR. lines makes the exterior angle equal to the in- the two straight lines AB, CD, make the exterior A G * angle EGB equal to the interior and opposite E angles on the same side BGH, GHD together equal B to two right angles : AB C -D shall be parallel to CD. H Because the angle EGB is equal t to the angle GHD, +Нур and the angle EGB equal*, to the angle AGH, *15. 1. therefore the angle AGH is equal t to the angle +1 Ax. GHD: and they are the alternate angles; therefore AB is parallel* to CD. Again, because the *27. 1. angles BGH, GHD are equal* to two right * Hyp. angles, and that AGH, BGH, are also equal * to *13. 1. two right angles; therefore the angles AGH, BGH are equalf to the angles BGH, GHD: f1 Ax Take away the common angle BGH; therefore the remaining angle AGH is equal † to the +3 Ax. remaining angle GHD: and they are alternate angles : therefore ABis parallelt to CD. Where- +27. 1. fore, if a straight line, &c. Q. E. D. PROP. XXIX. THEOR. If a straight line fall upon two parallel straight See the lines, it makes the alternate angles equal to one another; and the exterior angle equal to the position interior and opposite upon the sume side ; and Oct. Ed. 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, shall be equal to one another: and the exterior angle EGB shall be equal to the Interior and opposite, upon he same side, GHD; and the E two interior angles BGH, GHD upon the same side, A shall be together equal to two right angles. H D For, if AGH be not equal F notes on this pro on Oct. Ed. to GHD, one of them must be greater than the 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 +4 Ax. angles AGH, BGH are greater than theangles BGH, GHD: but the angles AGH, BGH, are *13. l. equal* to two right angles; therefore the angles BGH, GAD, are less than two right angles : but those straight lines which, with another straight line falling upon them, make the in terior angles on the same side less than two *12 Ax. right angles, will meet * together if continually Sereshen produced ; therefore the straight lines AB, CD), this pro- if produced far enough, will meet : but they position never meet, since they are parallel by the hypothesis ; therefore the angle AGH is not un equal to the angle GHD, that is, it is equal to *15. 1. it: but the angle AGH is equal * to the angle +1 Ax. EGB; therefore likewise EGB is equal t to GHD: add to each of these the angle BGH; +2 Ax. therefore the angles EGB, BGH are equal t to the angles BGH, GHD: but EGB, BGH are *13. 1. equal* to two right angles; therefore also BGH, *1 Ax. GHD are equal * to two right angles. Wherefore, if a straight line, &c. Q. E. D. PROP. XXX. THEOR. straight line are parallel to each other. Let the straight line GHK cut AB, EF, CD: and because GẮK cuts the parallel straight lines AB, EF, the angle AGH is * 29. 1. equal * to the angle GHF. Again, because the straight A -B line GK cuts the parallel Н straight lines EF, CD, the E -F *28. 1. angle GHF is equal* to the C D angle GKD: and it was shown that the angle AGK is equal to the angle GHF; therefore also AGK is equal * to GKD: and they are alternate +1 Ax. angles; therefore AB is parallel * to CD. * 27. 1. Wherefore, straight lines, &c. Q. E. D. PROP. XXXI. PROB. parallel to a given straight line. E A F In BC take any point D, and join AD; and at the point A, in the straight line AD, make* B 1) the angle DAE equal to the angle ADC; 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. Therefore the straight *27. 1. line EAF is drawn through the given point A, parallel to the given straight line BC. Which was to be done. C *23.1. * PROP. XXXII. THEOR. angle is equal to the two interior and opposite 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, viz. ABC, BCA, CAB, shall together be equal to two right angles. : Through the point C *31. 1. draw CE parallel* to the E straight line AB: and because AB is parallel to CE, and AC meets them, D the alternate angles BAC, * 29. 1. ACE, are equal.* Again, because AB is parallel to CE, and BD falls upon them, the exterior + 29. 1. angle ECD is equal t to the interior and opposite angle ABC : but the angle ACE was shown to be equal to the angle BAČ; therefore the whole +2 Ax., exterior angle ACD is equal to the two interior and opposite angles CAB, ABC: to each of these equals add the angle ACB, and the angles ACD, +2 Ax. ACB are equal t to the three angles CBA, BAC, * 13. 1. ACB; but the angles ACD, ACB are equal * to two right angles ; therefore also the angles CBA, +1 Ax. BAC, ACB, are equal t to two right angles. Wherefore, if a side of a triangle, &c. Q. E. D. Cor. 1.-All the interior an- For any rectilineal figure F, which is the common vertex of the triangles; *2 Cor. that is,* together with four right angles. There fore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides. Cor. 2.-All the exterior angles of any rectili. neal figure are together equal to four right angles. 15. 1. |