And if points in RS on one side of A are farther from BC than A is, so also are there corresponding points on the other side of A farther from BC than A is; which is impossible. .. all points in RS are equidistant from BC. Again, if any line be drawn through A other than RAS, and points taken along that part of it which is on the same side of RAS as BC is, farther and farther from A, their distances from RAS will at length be greater than AQ: (I. G.) .. the straight line thus drawn will at length cut BC. Hence through A no other straight line besides RAS can be drawn which shall never meet BC. DEF. Straight lines, lying in the same plane and which will never meet, though produced ever so far both ways, are called parallel. [The name being due to the property established in the foregoing Lemma.] PROPOSITION XX. If, in a plane, a straight line, meeting two other straight lines, makes the alternate angles equal to one another, these two straight lines shall be parallel. Let the straight line GH, meeting the two straight lines PQ, RS, make the alternate s PGH, GHS equal to one another. Then shall PQ be || to RS. For, if not, PQ, RS when produced will meet either towards QS or towards PR. Let them, if possible, be produced to meet towards Q, S in the point X. Then GHX is a ▲; and the exterior ▲ PGH is > the interior and opposite L GHX; but it is also equal to it, which is impossible. (I. 13) .. PQ, RS do not meet when produced towards Q, S. Similarly it may be proved that they will not meet if produced towards P, R. .. they are parallel. Hence are easily deduced the following Theorems : If a straight line, falling upon two other straight lines, makes the exterior angle equal to the interior and opposite upon the same side of the line, the two straight lines shall be parallel. Let the straight line EFGH, falling upon the two straight lines AB, CD, make the exterior angle EFB = the interior and opposite FGD upon the same side of EF. Then shall AB be || to CD. For the EFB is : = the vertical / AFG; = (1.6) .. LAFG is the ▲ FGD, and these are alternate angles ; .. AB is to CD. Also If a straight line falling upon two other straight lines makes the interior angles upon the same side of the line together equal to two right Ls, these two straight lines shall be parallel. For let the 4s BFG, FGD be together equal to rights; ... they are togethers BFG, AFG: take away the common ↳ BFG; (1.9) LFGD is = L AFG, and these are alternate angles; .. AB is I to CD. PROPOSITION XXI. If a straight line meets two parallel straight lines it makes the alternate angles equal to one another. Let the straight line FE meet the two || straight lines A AB, CD. Then shall ▲ AFG be = the alternate: 4 FGD. G For if not, if possible, at F in FG draw another line XFY, making 4 XFG = L FGD. Then XFY is | to CD. (I. 20) .. through Ftwo straight lines have been drawn to CD, which is impossible. .. LAFG is the alternate ▲ FGD. (Lemma.) Hence are easily deduced the following Theorems : If a straight line fall upon two parallel straight lines, it makes the exterior L = the interior and opposite upon the same side. L FGD. And AFG is the vertically opposite & HFB; (1.6) = .. & HFB is = 4 FGD. If a straight line fall upon two parallel straight lines, the two interior angles upon the same side shall together be equal to two right angles. That is, the 4S BFG, FGD shall be together = two right angles. and.. Ls AFG, BFG are together = 4 s FGD, BFG. ¿s But s AFG, BFG are together = two rights; (1.9) :. Ls FGD, BFG are together = two right 4 s. COR. If AB, BC are two straight lines respectively || to DE, EF, then shall ▲ ABC be = L DEF. G For, let BC, ED, produced, if necessary, intersect in H. Then 4 SABC, DEF are each = ▲ DHG, and are therefore equal to one another. * |