[I. 3. Next, let sides which are opposite to equal angles in each triangle be equal to one another, namely, AB to DE : likewise in this case the other sides shall be equal, each to each, namely, BC to EF, and AC to DF, and also the third angle BAC equal to the third angle EDF. For if BC be not equal to EF, one of them must be greater than the other. Let BC be the greater, and make BH equal to EF, and join AH. Then because BH is equal to EF, [Construction. and AB to DE; [Hypothesis. the two sides AB, BH are equal to the two sides DE, EF, each to each ; and the angle A BH is equal to the angle DEF ;[Hypothesis. therefore the triangle ABH is equal to the triangle DEF, and the other angles to the other angles, each to each, to which the equal sides are opposite ; [I. 4. therefore the angle BHA is equal to the angle EFD. But the angle EFD is equal to the angle BCA. [Hypothesis. Therefore the angle BHA is equal to the angle BCA; [Ax.1. that is, the exterior angle BHA of the triangle AHC is equal to its interior opposite angle BCA; which is impossible. [I. 16. Therefore BC is not unequal to EF, that is, it is equal to it; and AB is equal to DE; [Hypothesis. therefore the two sides AB, BC are equal to the two sides DE, EF, each to each ; and the angle ABC is equal to the angle DEF;[Hypothesis. therefore the base AC is equal to the base DF, and the third angle BAC to the third angle EDF. [I. 4. Wherefore, if two triangles &c. Q.E.D. PROPOSITION 27. THEOREM. If a straight line falling on two other straight lines, make the alternate angles equal to one another, the two straight lines shall be parallel to one another. Let the straight line EF, which falls on the two straight lines AB, CD, make the alternate angles AEF, EFD equal to one another : AB shall be parallel to CD. For if not, AB and CD, being produced, will meet either towards B, D or towards A, C. Let them be produced and meet towards B, D at the point G. Therefore GEF is a triangle, and its exterior angle AEF is greater than the interior opposite angle EFG; [1. 16. But the angle AEF is also equal to the angle EFG ; [Hyp. which is impossible. Therefore AB and CD being produced, do not meet towards B, D. In the same manner, it may be shewn that they do not meet towards A, C. But those straight lines which being produced ever so far both ways do not meet, are parallel. [Definition 35. Therefore AB is parallel to CD. Wherefore, if a straight line &c. Q.E.D. PROPOSITION 28. THEOREM. If a straight line falling on two other straight lines, make the exterior angle equal to the interior and opposite angle on the same side of the line, or make the interior angles on the same side together equal to two right angles, the two straight lines shall be parallel to one another. E Let the straight line EF, which falls on the_two straight lines AB, CD, make the exterior angle EGB equal to the interior and opposite angle GHD on the same side, or make the interior angles on the same side BGH, GHD together equal to two right angles : AB shall be parallel to CD. Because the angle EGB is equal to the angle GHD,[Hyp. and the angle EGB is also equal A G -B to the angle AGH, [I. 15. therefore the angle AGH is equal to the angle GHD; [Ax. 1. and they are alternate angles ; therefore AB is parallel to CD. [1. 27. Again; because the angles BGH, GHD are together equal to two right angles, [Hypothesis. and the angles AGH, BGH are also together equal to two right angles, [1. 13. therefore the angles AGH, BGH are equal to the angles BGH, GHD. Take away the common angle BGH; therefore the remaining angle AGH is equal to the remaining angle GHD; [Axiom 3. and they are alternate angles ; therefore AB is parallel to CD. [I. 27. Wherefore, if a straight line &c. Q.E.D. PROPOSITION 29. THEOREM. If a straight line fall on two parallel straight lines, it makes the alternate angles equal to one another, and the exterior angle equal to the interior and opposite angle on the same side ; and also the two interior angles on the same side together equal to two right angles. Let the straight line EF fall on the two 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 angle B on the same side, GHD, and the two interior angles ou the same side, BGH, GHD, shall be together equal to two right angles. For if the angle AGH be E not equal to the angle GHD, one of them must be greater than the other ; let the angle AGH be the greater. Then the angle AGH is greater D than the angle GHD; to each of them add the angle E BGH; therefore the angles AGH, BGH are greater than the angles BGH, GĦD. But the angles AGH, BGH are together equal to two right angles; [I. 13. therefore the angles BGH, GHD are together less than two right angles. But if a straight line meet two straight lines, so as to make the two interior angles on the same side of it, taken together, less than two right angles, these straight lines being continually produced, shall at length meet on that side on which are the angles which are less than two right angles. [Axiom 12. Therefore the straight lines AB, CD, if continually produced, will meet. But they never meet, since they are parallel by 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 to the angle EGB. [I. 15. Therefore the angle EGB is equal to the angle GHD. [Ax. 1. Add to each of these the angle BGH. Therefore the angles EGB, BGH are equal to the angles BGH, GHD. [Axiom 2. But the angles EGB, BGH are together equal to two right angles. [I. 13 Therefore the angles BGH, GHD are together equal to two right angles. [Axiom 1. Wherefore, if a straight line &c. Q.E.D. B Straight lines which are parallel to the same straight line are parallel to each other. Let AB, CD be each of them parallel to EF: AB shall be parallel to CD. Let the straight line GHK cut AB, EF, CD. Then, because GHK cuts the parallel straight lines AB, A EF, the angle AGH is equal to the angle GHF. [I. 29. E H/ Again, because GK cuts the parallel straight lines EF, C- K D CD, the angle GHF is equal to the angle GKD. [I. 29. And it was shewn that the angle AGK is equal to the angle GHF. Therefore the angle AGK is equal to the angle GKD;[Ax. 1. and they are alternate angles ; therefore AB is parallel to CD. [I. 27. Wherefore, straight lines &c. Q.E.D. PROPOSITION 31. 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 E D, and join AD; at the point A in the straight line AD, make the angle DAE equal to the alternate angle ADC; [I. 23. and produce the straight line EA to F. EF shall be parallel to BC. |