the less to the greater, which is impossible; therefore 5. AB is not unequal to DE, that is, it is equal to it; and BC is equal (Hyp.) to EF; therefore 6. The two AB, BC, are equal to the two DE, EF, each to each; and the angle ABC is equal (Hyp.) to the angle DEF; therefore (I. 4.) 7. The base AC is equal to the base DF, and the third angle BAC to the third angle EDF. Next, let the sides which are opposite to equal angles in each triangle be equal to one another, viz. AB to DE; likewise in this case, the other sides shall be equal, AC to DF, and BC to EF; and also the third angle BAC to the third EDF. 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, 1. to each; The two AB, BH, are equal to the two DE, EF, each and they contain equal angles; therefore (I. 4.) 2. The base AH is equal to the base DF, and the triangle ABH to the triangle DEF, and the other angles to the other angles, each to each, to which the equal sides are opposite; therefore 3. The angle BHA is equal to the angle EFD; but EFD is equal (Hyp.) to the angle BCA; therefore also 4. 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 (I. 16.); wherefore BC is not unequal to EF, 5. that is, it is equal to it; and AB is equal (Hyp.) to DE; therefore 6. The two AB, BC, are equal to the two DE, EF, each to each; and they contain (Hyp.) equal angles; wherefore (I. 4.) 7. 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. PROP. XXVII.—THEOREM. If a straight line, falling upon two other straight lines, make 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 another; 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 1. GEF is a triangle, and its exterior angle AEF is greater (I. 16.) than the interior and opposite angle EFG; but it is also equal to it, which is impossible; therefore 2. AB and CD being produced do not meet towards B, D. In like manner it may be demonstrated that 3. AB and CD do not meet towards A, C; but those straight lines which meet neither way, though produced ever so far, are parallel (Def. 35.) to one another; therefore 4. AB is parallel to CD. Wherefore if a straight line, &c. Q.E.D. PROP. XXVIII.—THEOREM. If a straight line falling upon two other straight lines make the exterior angle equal to the interior and opposite upon the same side of the line; or make the interior angles upon the same side together equal to two right angles; the two straight lines shall be parallel to one another. 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 (I. 15.) to the angle AGH, 1. The angle AGH is equal to the angle GHD, and they are alternate angles; therefore (I. 27.) 2. AB is parallel to CD. Again, because the angles BGH, GHD, are equal (Hyp.) to two right angles, and that AGH, BGH, are also equal (I. 13.) to two right angles, 1. The angles AGH, BGH, are equal to the angles BGH, GHD. Take away the common angle BGH; therefore 2. The remaining angle AGH is equal to the remaining angle GHD, and they are alternate angles; therefore (I. 27.) 3. AB is parallel to CD. Wherefore if a straight line, &c. Q.E.D. If a straight line full 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. For if AGH be not equal 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 1. The angles AGH, BGH, are greater than the angles BGH, GHD; but the angles AGH, BGH, are equal (I. 13,) to two right angles; therefore 2. 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, will meet (Ax. 12.) together if continually produced; therefore 3. The straight lines AB, CD, shall meet, if produced far enough; but they never meet, since they are parallel by the hypothesis; therefore 4. The angle AGH is not unequal to the angle GHD, that is, it is equal to it; but (I. 15.) the angle AGH is equal to the angle EGB; therefore likewise 5. EGB is equal to GHD; add to each of these the angle BGH; therefore 6. The angles EGB, BGH, are equal to the angles BGH, GHD; but EGB, BGH, are equal (I. 13.) to two right angles; therefore also 7. BGH, GHD, are equal to two right angles. Wherefore, if a straight line, &c. Q.E.D. PROP. 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, CD; and because GHK cuts the parallel straight lines AB, EF, therefore (I. 29.) 1. The angle AGH is equal to the angle GHF. Again, because the straight line GHK cuts the parallel straight lines EF, CD, (I. 29.) 2. The angle GHF is equal to the angle GKD; and it was shown that the angle AGK is equal to the angle GHF; therefore also 3. AGK is equal to GKD; and they are alternate angles; therefore (I. 27.) 4. AB is parallel to CD. Wherefore straight lines, &c. Q.E.D. PROP. 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 4, in the straight line AD, make (I. 23.) the angle DAE equal to the angle ADC; and produce the straight line EA to F; EF is 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, (I. 27.) EF is parallel 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. PROP. 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 D; the exterior angle ACD is equal to the two interior and opposite angles CAB, ABC; and the three interior angles of the triangles, viz. ABC, BCA, CAB, are together equal to two right angles. Through the point C draw CE parallel (I. 31.) to the straight line AB; and because AB is parallel to CE, and AC meets them, (I. 29.) 1. The alternate angles BAC, ACE, are equal. Again, because AB is parallel to CE, and BD falls upon them, 2. The exterior angle ECD is equal to the interior and opposite angle ABC; but the angle ACE was shown to be equal to the angle BAC; therefore (Ax. 2.) 3. The whole exterior angle ACD is equal to the two interior and opposite angles CAB, ABC; |