PROPOSITION XXV. THEOREM.--If two triangles have two sides of the one, equal to two sides of the other respectively, but the third side of the one is greater than the third side of the other; the angle between the sides of that which has its third side the greater, shall be greater than the angle between the sides equal to them, of the other. AN CE * 1. 4. Let ABC, DEF be two triangles which have B For, if it be not greater, it must either be equal or less. But the angle A cannot be equal to the angle D; because then the third side BC would be* equal to the third side EF; and it is not equal, for it ist greater. Neither can it be less ; because then the third side BC would be less than the third side EF; and it is not less, for it is greater. But because the angle A is neither equal to the angle D, nor less, it is greater. And by parity of reasoning, the like may be proved in all other triangles under the same conditions. Wherefore, universally, if two triangles have two sides of the one, &c. Which was to be demonstrated. + Hyp. I 1. 24. PROPOSITION XXVI. THEOREM.-If two triangles have two angles of the one, equal to two angles of the other respectively; and have moreover one side of the one triangle equal to one side of the other triangle, viz. either the sides which lie between the equal angles, or sides opposite to equal angles in each ; then shall the remaining angle of one triangle be equal to the remaining angle of the other; and the two triangles shall be equal ; and their other sides shall be equal respectively, viz. those which are opposite to equal angles. Let ABC, DEF be two triangles which have the angles ABC, ACB equal to the angles DEF, DFE respectively, viz. ABC to First Case ; where the side BC For, if BA be not equal to ED, one of them must be the * 1. 3. greater. Let BA be assumed to be the greater; and make* BG equal to ED, and join CG. + Hyp. Because BC ist equal to EF, and BG to ED, BC and BG are | Нур. equal to EF and ED respectively. Also the angle CBG is equal * 1.4. to the angle FED; therefore the triangles CBG, FED must be* equal, and the angle GCB, which is opposite to BG, must be equal to the angle DFE which is opposite to ED. But the angle DFE + Hyp. ist equal to the angle ACB; therefore the angle GCB (which is INTERC. 1. equal to DFE) must bef equal to ACB, the less to the greater, which is impossible. It cannot be, therefore, that BA is greater they are equal. Wherefore, BA being shown to be equal to ED, Нур. : and BC being* equal to EF, BA and BC are equal to ED and EF + Hyp. respectively; and also the angle ABC ist equal to the angle DEF; 1. 4. therefore AC isf equal to DF, and the triangle ABC to the tri- D B Cor. 1. * For if BC be not equal to EF, let BC be assumed to be the • 1.3. greater ; and make* BH equal to EF, and join AH. +Нур. . Because BH is equal to EF, and BA tot ED, BH and BA | Hyp. are equal to EF and ED respectively. Also the angle ABH is equal to the angle DEF; therefore the triangles ABH, DEF *1, 4. must be* equal, and the angle AHB which is opposite to BA must be equal to the angle DFE which is opposite to ED. But the + Hyp. angle DFE ist equal to the angle ACB or ACH; therefore the IINTERC. 1. Cor. 1. angle AHB (which was shown to be equal to DFE) must be equal to ACH, the exterior angle equal to the interior and opI. 16. posite, which is* impossible. It cannot be, therefore, that BC is greater than EF; and in the same manner may be shown that EF cannot be the greater; and because neither of them is greater than the other, they are equal. Wherefore, BC being | Hyp. shown equal to EF, and BA beingt equal to ED, BC and BA are Hyp. equal to EF and ED respectively, and also the angle ABC ist • 1. 4. equal to the angle DEF; therefore AC is* equal to DF, and the triangle ABC to the triangle DEF, and the remaining angle See Note. BAC to the remaining angle EDF. And by parity of reasoning, the like may be proved in all other triangles under the same conditions. Wherefore, universally, if two triangles have two angles of the one, equal to two angles of the other respectively ; &c. Which was to be demonstrated. PROPOSITION XXVII. THEOREM.-If a straight line falling upon two other straight lines makes the alternate angles equal to one another, those two straight lines being prolonged ever so far both ways, shall not meet. See Note. G First Case ; where the two straight lines E are both in the same plane. Let the straight А B line EF, falling upon the two straight lines AB, CD, which are both in the same plane, С H D F make the alternate angles GHC, HGB, equal to one another. AB and CD, being prolonged ever so far both ways, shall not meet. G For if this be disputed, let it be assumed that on being prolonged they meet on the side of B and D, in the point I. • I. 16. + Нур. . Therefore GHI must be a triangle, and its exterior angle GHC greater than the interior opposite angle HGI. Which is impossible, for it ist equal to it. It cannot be, therefore, that AB and CD being prolonged, will meet on the side of B and D. And in the same manner may be shown that they cannot meet on the side of A and C. So also if HGA, GHD had been the alternate angles taken to be equal. Second Case ; if the two straight lines are not in the same plane, they can never meet though prolonged ever so far both ways, whether the alternate angles they make with the other straight line are equal or not. IINTERC.14. For if they ever meet, they must lie wholly in one plane. Cor. 2. But they are not in one plane; therefore it cannot be true that they ever meet. And by parity of reasoning, the like may, in either of the Cases, be proved of all other straight lines under the same conditions. Wherefore, universally, if a straight line falling upon two other straight lines &c. Which was to be demonstrated. NOMENCLATURE.Straight lines which are in the same plane, and which being prolonged 'ever so far both ways do not meet, are called parallel. CoR. 1. If a straight line falling upon two other straight lines which are both in the same plane, makes the alternate angles equal to one another, those two straight lines shall be parallel. For it has been shown in the Proposition above, that being prolonged ever so far both ways they will not meet. COR. 2. If a straight line falling upon two other straight lines which are both in the same plane, makes the exterior angle equal to the interior and opposite on the same side of the line; or if it makes the two interior angles on the same side, together equal to two right angles; those two straight lines shall be parallel. G F Cor. 1. For First ; where the exterior angle EGA E is equal to the interior and opposite angle A B GHC. Because the angles EGA, HGB * 1.15. are vertical angles, they are* equal to one с H D + Hyp. another. And because the angle EGA ist INTERC. 1. equal to GHC, the angle HGB (which is equal to EGA) ist equal to GHC. And they are alternate angles; therefore (by Cor. 1 above) AB and CD are parallel. Secondly; where HGB and GHD the interior angles on the same side of the line, are together equal to two right angles. * Hyp: Because the angles HGB, GHD are together* equal to two right + I. 12. angles, and the angles GHC, GHD are also togethert equal to INTERC. 1. two right angles; the angles HGB, GHD are together equal to GHC, GHD. Take away the common angle GHD; therefore *INTERC. 1. the remaining angle HGB is* equal to the remaining angle GHC. Cor. 7. And they are alternate angles; therefore AB and CD (by Cor. 1 above) are parallel. COR. 3. If the exterior angle EGA is less than the interior and opposite angle GHC, GA and HC being prolonged ever so far, will not meet on the side of A and C. For (by Cor. 2 above) when the angle GHC is equal to EGA, GA being prolonged ever so far will not meet HC. Still less can it meet a straight line drawn from H and lying on the other side of HC; as will be the straight line that makes with HG an angle greater than GHC or than its equal EGA. PROPOSITION XXVIII. See Note. PROBLEM.-Through an assigned point to draw a straight line, parallel to an assigned straight line. A F Let A be the assigned point, and BC the E с |