than the side EF; but Eg is equal to BC; and therefore also b c is greater than E F. Therefore, if two triangles, &c. Q. E. D. D A PROP. XXV. THEOR. If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of the one greater than the base of the other; the angle also contained by the sides of that which has the greater base, shall be greater than the angle contained by the sides equal to them, of the other. Let A BC, D E F be two triangles which have the two sides A B, AC equal to the two sides DE, DF, each to each, viz. A B equal to D Е, and AC to DF; but the base cB is greater than the base E F; the angle B A C is likewise greater than the angle E D F. For, if it be not greater, it must either be equal to it, or less; but the angle BAC is not equal to the angle E DF, because then the base BC would be equal (1. 4) to EF; but it is not, therefore the angle BAC is not equal to the angle EDF; neither is it less; be E cause then the base BC would be less (1. 24.) than the base E F; but it is not; therefore the angle BAC is not less than the angle EDF; and it was shown that it is not equal to it; therefore the angle Bac is greater than the angle EDF. Wherefore if two triangles, &c. Q. E. D. PROP. XXVI. THEOR. If two triangles have two angles of one equal to two angles of the other, each to each ; and one side B F D B E F equal to one side, viz. either the sides adjacent to the equal angles, or the sides opposite to equal angles in each ; then shall the other sides be equal each to each : and also the third angle of the one to the third angle of the other. Let A B C D E F be two triangles which have the angles A B C, BCA equal to the angles D E F, EF D, viz. ABC to D E F, and BC A LO E FD; also one side equal to one side ; and first let those sides be equal which are adjacent to the A angles that are equal in the two triangles, viz. BC to EF; the other sides shall be equal, each to each, viz. A B to DE, and AC to DF; and the third angle BAC to the third angle E D F. For, if A B be not equal to D Е, one of them must be the greater. Let A B be the greater of the two, and make B G equal to D Е, and join GC; therefore, because BG is equal to D Е, and B C to E F (Hyp.), the two sides G B, BC are equal to the two DE, E F, each to each ; and the angle GBC is equal to the angle D E F; therefore the base Gc is equal (1. 4.) to the base DF, and the triangle GBC to the triangle D E F, and the other angles to the other angles, each to each, to which the equal sides are opposite; therefore the angle G C B is equal to the angle D FE; but DFE is, by the hypothesis, equal to the angle BCA; wherefore also the angle BC G is equal to the angle B C A, the less to the greater, which is impossible ; therefore A B is not unequal to D Е, that is, it is equal to it; and Bc is equal to E F : therefore the two A B, BC are equal to the two DE, E F, each to each ; and the angle ABC is equal to the angle DEF; the base therefore ac is equal (1. 4.) to the D base DF, and the third angle BAC to the third angle EDF. Next, let the sides which A are opposite to equal angles in each triangle be equal to one another, viz. A B to DE; likewise in this case, be BC B to EF; and also the third angle BAC to the third HC E EDF. For, if Bc be not equal to E F, let Bc be the greater of them, and make bh equal to E F, and join Ah; and because Bh is equal to E F, and A B to DE (Hyp.), the two A B, BH are equal to the two DE, E F, each to each ; and they contain equal angles (Hyp.); therefore the base a H is equal to the base D F, and the triangle A B H to the triangle DE F, and the other angles shall be equal, each to each, to which the equal sides are opposite; therefore the angle BH A is equal to the angle E FD; but EFD is equal to the angle BCA (Hyp.); 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 (1. 16.); wherefore Bc is not unequal to E F, that is, it is equal to it; and A B is equal to DE (Hyp.); therefore the two A B, BC are equal to the two DE, E F, each to each; and they contain equal angles ; wherefore the base A c is equal to the base D F, and the third angle BAC to the third angle E D F. Therefore if two triangles, &c. Q. E. D. D 3 PROP. XXVII. THEOR. 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 É F, which falls upon the two straight lines A B, C D, make the alternate angles A E F, E F D equal to one another; A B is parallel to CD. or B For, if it be not parallel, A B and cd being produced, shall meet either towards B, D, towards A, C; let them be produced and meet towards B, D in the point g; therefore G EF is a triangle, and its exterior angle A E F is greater (1. 16.) than the interior and opposite angle EFG; A E but it is also equal to it (Hyp.), which is c impossible: therefore A B and cd being produced do not meet towards B, D. In like manner it may be demonstrated that they do not meet towards AC; but those straight lines which meet neither way, though produced ever so far, are parallel (Def. 35.) to one another. AB therefore is parallel to CD. Wherefore if a straight line, &c. Q. E. D. D PROP. XXVIII. THEOR. If a straight line falling upon two other straightlines 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. E upon the A B Let the straight line E F, which falls two straight lines A B, CD, make the exterior angle E G B equal to the interior and opposite angle G H D upon the same side; or make the interior angles on the same side BGH, GHD together equal to two right angles; A B is parallel to с D 11 F CD. Because the angle E GB is equal to the angle G I D, and the angle E G B equal (1. 15.) to the angle A GH, the angle A G H is equal to the angle G HD, and they are the alternate angles; therefore A B is parallel (1. 27.) to CD. Again, because the angles BGH, GHD are equal (by Hyp.) to two right angles, and that A G H, B G H are also equal (1. 13.) to two right angles, the angles, A GH, B G H are equal to the angles B G H, GHD: Take away the common angle BGH; therefore the remaining angle A G H is equal to the remaining angle G A D, and they are alternate angles; therefore AB is parallel to CD (1. 27.). Wherefore if a straight line, &c. Q. E. D. PROP. XXIX. THEOR. 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 |