PROPOSITION XXXIII. Equal triangles on the same base and on the same side of it are between the same parallels. Let FGH, PGH be equal as upon the same base GH, and upon the same side of it. Join FP. Then shall FP be || to GH. For if not through F draw FQ || to GH, meeting PG or PG produced in Q, and join QH. ... the temporary hypothesis is false, which was that FP is not || to GH. ... FP is || to GH. COR. By a similar proof it may be shewn that equal triangles upon equal bases in the same straight line and on the same side of that line are between the same parallels. PROPOSITION XXXIV. The complements of the parallelograms, which are about the diameter of any parallelogram, are equal to one another. Let FGHK be a , and GK one of its diagonals, and SR, PN 's about the diagonal. Then FQ, QH, the remaining ☐s, which are called complements, shall be equal to one another. For ... s are bisected by their diagonals, (1. 26) .. △ SGQ is = RGQ, and △PQK = △NQK; .. AS SGQ, PQK are together = △s RGQ, NQK; but the whole △ FGK is = the △ HGK; ... the remainder FQ is = the remainder QH. PROPOSITION XXXV. In any right-angled triangle, the square described on the side opposite the right angle is equal to the squares described on the sides containing the right angle. is = the right & QPH; add to each the RPQ; ... the 2 SPQ is = the K Hence SP, PQ and the 2 SPQ are respectively = RP, RPH. are on the same base and between the same parallels. (1.32) Also. FRP and QRP are right ∠s; ... FRQ is a straight line (1. 10), and it is || to SP. ... □PC is = SPRF the square on PR. Similarly it may be proved that the square on RQ; (Ι. 32) QC is = the ... the whole figure PHNQ, that is the square on PQ, is = the squares on PR, RQ. PROPOSITION XXXVI. If the square described on one of the sides of a triangle be equal to the squares described on the other two sides of it, the angle contained by those sides is a right angle. R Let the square described on the side PQ of the △ PQR be equal to the squares on PR, RQ. Then shall the PRQ be a right 2. Through R draw RS 1 to PR and = RQ. Now. SR is = QR; ... square on SR is = square on QR. ... the squares on SR, RP are = the squares on QR, RP, but the squares on SR, RP are together = the square on SP; (1. 35) ... SRP is a right ; and the squares on QR, RP are = the square on QP; (hyp.) ... the square on SP is = the square on QP; ... SP is = QP. Hence, PR, RS, SP are respectively = PR, RQ, QP. PROBLEM L. Describe a parallelogram equal to a given triangle, and having an angle equal to a given angle. Let FGH be the given, and K the given angle. It is required to describe a = ▲ FGH and having an angle = K. Bisect GH in R; join FR; through F draw FST || to GH; at R in RH make HRS = K. Through Hdraw HT || to RS; then SRHT shall be the required. • GR is = RH; .. △ FGR is = △ FRH; (Ι. 32) .. AFGH is double of the △ FRH; also the SH is double of FRH, they are upon the same base and between the same parallels. (Ι. 32) SH is = △ FGH, and it has an SRH equal to the given angle K. |