equal to the triangles DBE, GIC; .. the triangle ABC is equal to AEFG, i. e. to the given figure. (39.) Between two lines given in position, to draw a line equal to a given line, so that the triangle thus formed may be equal to a given rectilineal figure. Let AB, AC be the lines given in position, and DE the line whose magnitude is given. Bisect it in F, and on DF describe a rectangular parallelo H G DA K B gram equal to the given figure. On DE describe a segment of a circle containing an angle equal to the angle at A, and cutting HG in I. Join DI, IE; and make AK ID, and AL-IE. Join KL; it is the line required. Since AK= ID, and AL= IE, and the angle at A= DIE, :. KL=DE, and the triangle AKL=IDE= HGFD=the given figure. (40.) From two given lines to cut off two others, so that the remainder of one may have to the part cut off from the other a given ratio; and the difference of the squares of the other remainder and part cut off from the first may be equal to a given square. Perpendicular to AB one of the given lines, draw BC equal to a side of the given square; and take AD to the other given line in the given ratio of the part remaining from the first to the part cut off from the BG and GF are equal to the parts to be cut off. B G For the difference between the squares of CG, GB is equal to the square of BC, i. e. to the given square; and AG : GF :: AD : AE, i, e. in the given ratio. (41.) From two given lines to cut off two others which shall have a given ratio, so that the difference of the squares of the remainders may be equal to a given square. Let AC be one of the two given lines. B From C draw CD perpendicular to AC, and equal to a side of the given square. Take AE to the other given line in the given ratio of the parts to be cut off. Join ED, and produce it; and with the centre A, and radius equal to that other given line, describe a circle cutting ED in B. Join AB; and let it meet DF, which is parallel to AC, in F. Draw FG parallel to CD. CG and BF are the parts required to be cut off. For (DF =) CG : FB :: EA : AB,'i. e. in the given ratio of the parts to be cut off, and the difference between the squares of FA and AG is equal to the square of GF, i. e. to the square of CD, or the given difference of the squares of the remainders. (42.) From two given lines to cut off two others so that the remainders may have a given ratio, and the sum of the squares of the parts cut off may be equal to the square of a given line. Let AB be one of the given lines, and in it take AC to the other given line, in the given ratio of the remainders. From Cdraw CD perpendicular to AB, and equal to the second given line. Join B AD, and draw CE parallel to AD; and with the centre B, and radius equal to the side of the given square, describe a circle, cutting CE in E.. Draw EF parallel to DC. Then BG, GE will be equal to the parts to be cut off. Join BE. The squares of BG, GE are equal to the square of BE, i. e. to the given square; and AG: GF :: AC: CD, i. e. in the given ratio of the remainders. (43.) Two points being given in a given straight line; to determine a third such that the rectangles contained by its distances from each extremity and the given point adjacent to that extremity may be equal. Let AB be the given straight line, C and D the given points in it. On AC and DB as dia meters let circles be described, and let EF touch them in E A E N B CH D M and F. Bisect EF in G, and let fall the perpendicular GH; H is the point required. From G draw any lines GNK, GLM cutting the circles. Take O the centre of the circle ACE, and draw OP perpendicular to GK. (Eucl. ii. 6.) the rectangle NG, GK together with the square of PN is equal to the square of PG; to each of these add the square of PO; and the rectangle NG, GK together with the squares of OP, PN (i. e. the square of OC) is equal to the squares of OP, PG, i. e. to the square of OG, or to the squares of OH, HG. But the square of OH is equal to the rectangle CH, HA together with the square of OC; whence the rectangle NG, GK is equal to the rectangle CH, HA together with the square of HG. In the same manner it may be shewn that the rectangle LG, GM is equal to the rectangle DH, HB together with the square of HG. But since the rectangle NG, GK, is equal to the rectangle LG, GM, the rectangle CH, HA is equal to the rectangle DH, HB. COR. If IH be a mean proportional between CH and HA; IG=GE. (44.) Through the point of intersection of two given circles, to draw a line in such a manner that the sum of the respective rectangles contained by the parts thereof which are intercepted between the said point and their circumferences, and given lines A and B, may be equal to a given square. Let the two circles CID, CEK cut each other in the point C; from C draw the diameters CD, CE. In CD take the point Fsuch, that CD: CF: A B. Join EF; and on E K it as a diameter describe a semicircle, in which place EG a third proportional to A and the side of the given square. Draw ICK parallel to EG; it will be the line required. Join FG, and produce it to H. The angle DIC is equal to FGE, i. e. to FHC, .. FH is parallel to DI; and CI CH:: CD CF :: A: B, .. the rectangle A, CH, is equal to the rectangle B, CI. Now since EG is a third proportional to A and the side of the given square, the rectangle A, EG will be equal to the given square. But the rectangle A, EG, is equal to the rectangles A, HC, and A, CK, i, e. to the rectangles B, IC, and A, CK; . the rectangles A, KC, and B, IC, are equal to the given square. (45.) Through a given point, to draw an indefinite line, such, that if lines be drawn from two other given points, and forming given angles with it, the rectangle contained by the segments intercepted between the given point and the two lines so drawn, shall be equal to the square of a given line. Let A be the given point through which the line is to be drawn; B and C the other given points. Join AB, AC; and on them describe segments of circles ADB, AEC, containing G angles equal to the given angles. Draw either diameter AF, on which produced take AG such, that the rectangle FA, AG, may be equal to the given square. Draw GE perpendicular to GF; join EA, and produce it both ways; it is the line required. Join DF. The angles at G and D being right angles, the triangles AGE, ADF are similar, .. EA: AG :: FA: AD, ..the rectangle EA, AD is equal to the rectangle FA, |