Conftru&tion. On the longest Leg AB of the given Triangle defcribe a Semicira cle, and in it apply AC the other E Leg, drawing alfo CB; fo will CBg: beddcc: Then on CB defcribe another Semicircle, and therein apply CB the Bafe of the Triangle given, drawing liked Wife The Line B6: Then will & Bq be — dd — cc - bb And confequently & B = √ dd - cc - 66: And lince the firft part of the eighth Step is divided by 2b, make as BE = 2 b .. BB = d2 = c2 = b2 | 1⁄2 : : B F = BB. ¿D=DC the Segment fought CHA P. II. A Problem producing a Simple Quadratick Equation. Having given the Square B AD, and a Right Line N BN, you are to produce the It will be evident, if you imagine a Semicircle to paß thro the Points B and E, that the most commodious way will be to find B C N H the Line DG, that you may have the Diameter BG, upon which having afterwards defcrib'd a Semicircle, there will be need of no other Operation to fatisfy the Problem, than to produce the Side AC till it occur the preferib'& Periphery. Solu the ABFD, GEH and 7b (BD) y (B F) : : b (EH) •• y (E G.) BEG are fimilar 8b+x(BG)..y (EG): :j+e(BE) ..b (EH) Equation 9 bb + bx=yy+yc. 47. 1. Eucl. El. 10+2 6 x + x x (BG q) = yy (E Gq :) {bb 2b xx +y+246 + cc (BE q :) 11bx+xx=yy + yo+cc. 10 9 - cc 12 bx + x x 12=9 13 bx+xxcc (= y y + y c) = b b + b si 13-bx + c2 14 x x = b b + co. 14 w 12/15│x=√bb+cc=5. = Conftruction. Having produc'd the Side of the Square BA to N, so that BN fhall be the given Right Line BN; then fince BD is b, and B Nc, the Hypothenufe DN will be = bb + c c√ z = x : Having therefore made DG DN, and defcrib'd a Semicircle upon the whole Line B G, if AC be prolong'd until it occur the Periphery in E, you'll have done that which was requir'd. CHA P. III. CHA P. III. Of PROBLEMS producing Adfected Quadratick Equations. PROBLEM I. N the LA ABC, the Per Ipendicular, BC, and the alter nate Segment of the Hypothenuse, (made by a 1 let fall from the L) viz. A D being given, to find the other Segment D C, &c. Solution. Let IBC=p the I= 45 Andfuppofe 3 a DC By 4. 6. = 46. BD BD ....a Aq. By 47. I. Eucl. El. aa S — = 5= 6 7 ba 7 + aa 8 a2 + ba = pp. Cafe 1. Of Adfected Qua dratick Equations. Comp. 9a+ba + 2 bb = pp + 4 bb 90210 a + b = √ pp + 4 bb 10 — 1611a: = √ pp + & bb - 1 b = 27. Points of the; then compleat the Diameter, by produBbb cing Part II. cing A D to O and C; then will DC, or O A, be = a for OD (a+b).. DL (p) :: DL (p). DC=a per 13. 6 Eucl. El. confequently aa + ba = pp; as in the 8th Step. PROBLEM II. The Difference A E between the Bafe A B and Perpendicular BC of the LABA C, and the Perpendicular BD let fall from the right Angle ABC upon the Hypothenufe A C being given, thence to B find the Hypothenufe AC. fuppofe 3a= and 4 e 4. 6: Eucl. Sd+c (AB). p (BD) a (AC) .. El. 352 BC= e 47.1.Encl.7 5 dd + 2de + ee El. 7 dd + 2de + 6 x 28 2de + 2ee = 2pa 9 ee (ABq:+BCq:) 2ee = aa ོ༡ -- 8 dd 10211 V dd + pp =a-p 10 + pl2p + v dd + BP = 9 = 75 N M With L A 2p, and A E d, makethed LAEN, and upon M, the middle Point of L A, defcribe fuch a Semicircle as will pafs thro' N and E, the emoteft Points of the then compleat the Diame ter, by producing LA to O and C; and CA LO will bea for CL (a-2p). LN (d): LN (d) a, per 13. 6. Eucl. El. Whence aa — ~.LO PROBLEM III. The Hypothenufe A C of any d AA BC, and the Perpendicular BD, let fall from the LA BC, upon the Hypothenufe AC being gi ven, to find A D, the greater Seg ment of the Hypothenufe. Solution. 2pa=dd; D 1/2 1/2 4 IOW211 a 11 + 1/2 b 12 a = bb PP = Cafe 3: 4 bb b 1 ± √ 4 bb — pp = 48 or 27 The Geometrical Conftruction of Cafe 3. viz. ha a app, may be thus perform'd. Draw a right Line of any convenient Length, as M Perpendicular fet off BMb; and upon the Point M, where B M touches A Z with the Radius M B, defcribe a Semicircle A BCM; then will its Diameter A C =b be cut by the Perpendicular B D into two Segments, AD and DC, which are the two Values of the Root a, B b b 2 viz. |