PROBLEM C. To describe an isosceles triangle having each of the angles at the base double of the angle at the vertex. Take any straight line AB and From Q let fall QP1 to BC, and .. bisecting BC. Then squares on AB, BQ are together > square on AQ by twice rectangle AB, BP. (II. 4) But rect. AB, BC = rect. AB, BP + rect. AB, PC and twice the rect. AB, BP; (II. I) .. square on AC = twice rectangle AB, BP; ... squares on AB, BQ are together squares on AQ, AC, and is.. double of CAQ. Thus each of the angles at the base BQ of ▲ ABQ is double of the angle at the vertex A. The Theorems in Proposition II. may be established independently as follows: THEOREM (a). If a straight line be divided into any two parts, the square on the whole line is equal to the squares on the parts together with twice the rectangle contained by the parts. Let the straight line PQ be divided into any two parts in C, then shall square on PQ be PC, CQ+ twice the rectangle PC, CQ. the squares on draw HM, RS || to PQ; and CF, QS || to PR. Then PRSQ is a square, and it is the figures PK, KS, CM, HF. KS is = square on CQ, for KM is = CQ; CM is rectangle PC, CQ, for CK is = PC, (I. 28) (1. 28) (1. 28) and HF is rectangle PC, CQ, for HK is = PC, and HR is = CQ; = .. square on PQ is squares on PC, CQ + twice rectangle PC, CQ. THEOREM (b). If a straight line be divided into any two parts, the square, on one part is less than the squares on the whole and on the other part by twice the rectangle contained by the whole and that other part. Let the straight line AB be divided into any two parts in C. Then shall the square on AC be < the squares on AB and BC by twice the rectangle AB, BC. Draw AGSL to AB, and QCH, RBT to AGS; draw GHF, ST and RQ || to AB. Then AH, AT and BQ are squares, and we have to shew that AH is < AT and BQ together by twice rectangle AB, BC, i.e. that the figure GTQ is twice rectangle AB, BC. = Now figure GT = rectangle AB, BC, for GF AB and GS = BC; = .. figure GTQ-twice rectangle AB, BC; (I. 26) ... square on AC is < squares on AB, BC by twice rectangle AB, BC. 1 THEOREM (c). The difference between the squares on any two straight lines is equal to the rectangle contained by two straight lines, one of which is equal to the sum, and the other equal to the difference of the given lines. Let RQ, RS be equal to the P given straight lines. Then SQ is = the difference between them. H R The difference between the squares on RQ, RS shall be rectangle (PS, SQ). On RQ describe a square, draw PH1 to PQ and = SQ; draw SF to RG and HKN || to PQ. = Then KF is square on RS, for KN is = RS; difference between the squares on RQ, RS is = figure KQF. We have now to shew that figure KQF-rectangle (PS, SQ). The rectangle QF= rectangle PK, for QS, QC are respectively PH, PR ; = ... figure KQF= figure PN, which is rectangle PS, SQ, for PH=SQ; .. the difference between the squares on RQ, RS = rect angle (PS, SQ). R R COR. P If PQ be bisected in R, and any point S be taken in PQ or PQ produced, then rectangle PS, SQ= difference between the squares on RQ, RS. THEOREM (d). The square on the sum of two straight lines is greater than the square on their difference by four times the rectangle contained by them. Let the two straight lines AB, BC be placed in the same straight line : cut off BD = BC. Then shall the square on AC be greater than the square on AD by four times the rectangle (AB, BC). DB On AC, AD describe squares, produce QR, DR: cut off QS BC, draw SW | to AC and BG || to AH. The square on AC is greater than the square on AD by figure DKQ. Hence we have to prove that figure DKQ=four times the rectangle (AB, BC). Now the rectangles SF, SR, DX, CX are equal to one another, also the squares FT, TK, RT, TY are equal to one another; ... figure DKQ = four times the figures SF, FT together; i. e. four times rectangle (HG, HS), and.. four times rectangle (AB, BC). = square on AC is greater than square on AD by four times rectangle (AB, BC). |