PROPOSITION XXXIV. The complements of the parallelograms, which are about the diameter of any parallelogram, are equal to one another. Let FGHK be a. D., and GK one of its diagonals, and SR, PN os about the diagonal. Then FQ, QH, the remaining s, which are called complements, shall be equal to one another. 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. Let PN, PF, RK be squares described on the sides of the right-angled - PQR. .. Then shall PN the square on the side opposite the right 2 be equal to PF, RK Now the right 2 SPR T I ON is = the right - QPH; add to each the LRPQ; .. the < SPQ is = the LRPH. Hence SP, PQ and the 2 SPQ are respectively = RP, PH, and the į RPH; .. the .SPQ is = the - RPH. (1. 1) Now the o PC is double of the RPH, because they are on the same base and between the same parallels. (1. 32) Also :: FRP and QRP are right zs; i. O SPRF is double of the A SPQ, (1. 32) O PC is = SPRE the square on PR. Similarly it may be proved that the O QC is = the square on RQ; .. 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. Let the square described on the side P Q of the a PQR be equal to the squares on PR, RQ. Then shall the . PRQ be a right 2. PROBLEM L. Describe a parallelogram equal to a given triangle, and having an angle equal to a given angle. Let FGH be the given a, and K the given angle. It is required to describe a 0 = a FGH and having an angle = K. ::: GR is = RH;.. AFGR is = A FRH; (1. 32) : AFGH is double of the FRH; also the OSH is double of A FRH, •: they are upon the same base and between the same parallels. (1. 32) .:: SH is = A FGH, and it has an LSRH equal to the given angle K. PROBLEM M. To a given straight line apply a parallelogram equal to a given parallelogram. Let FG be the given straight line, and LH the given O. Produce GF, and cut off FM = HK, . and on FM describe D PF equal in all respects to LH. Through G draw GQ i to PM meeting PC produced in Q. Join QF and produce it to meet PM produced in S. Through S draw SVR || to MG or PQ meeting CF and QG produced in V and R. Then FR is a D, and it shall be the one required. For FR, PF are the complements about the diagonal QS of the O PR, and they are .:. = one another, (1. 34) but PF is = LH; COR. Hence to a given straight line may be applied a parallelogram equal to a given triangle and having an angle equal to a given angle. For a o LH may be drawn = the given a and having an 2 = the given 4, and the remainder of the construction is the same as in the proposition. |