Let ac be the proposed square, and put the side вc, or cd, Then, if the difference of BD and Bс be put thenuse BD will be = x + d. d, the hypo But since, as in the former problem. BC2 + CD2, or 2BC2 = BD2, we shall have Which equation, being resolved according to the rule laid down for quadratics in the preceding part of the work, gives d + d √2, Which is the value of the side вc, as was required. PROBLEM 3.—The diagonal of a rectangle ABCD, and the perimeter, or sum of all its four sides, being given, to find the sides. a Let the diagonal AC === d, half the perimeter AB + BC= α and the base BC= x; then will the altitude AB = a And since, as in the former problem, AB2 + BC2 OC. C Ac2, we shall have Which last equation, being resolved, as in the former instance, gives Where a must be taken greater than d and less than d√2. PROBLEM 4.-The base and perpendicular of any plane triangle ABC being given to find the side of its inscribed square. Let EG be the inscribed square; and put BC= and the side of the square EH or EF = x. have Then, because the triangles ABC, AEH, are similar, we shall AD: BC :: AI: EH, or p: b :: (p Whence, taking the products of the means and extremes, there will arise b + p Where b and p may be any numbers whatever, either whole or fractional. PROBLEM 5. Having the lengths of three perpendiculars, EF, EG, EH, drawn from a certain point E, within an equilateral triangle ABC, to its three sides, to determine the sides. Draw the perpendicular AD, and having joined EA, EB, and EC, put EF = a, EG = b, EH = c, and BD (which is BC) = x. Then, since AB, BC, or ca, are each by Euc. 1, 47, 2x, we shall have, x2) = √ 3x2 * √ 3. AD = √(AB And because the area of any plane triangle is equal to half the rectangle of its base and perpendicular, it follows that A ABC A BEC A AEC A AEB 1 BC X AD = x × ∞ √ 3 = x2 √ 3, X x ABX EH x X c = ax, bx, = cx. But the last three triangles BEC, AEC, AEB, are together equal to the whole triangle ABC; whence x2 √3 = ax + bx + cx, And consequently. if each side of this equation be divided by x, we shall have Which is, therefore, half the length of either of the three equal sides of the triangle. COR.-Since, from what is above shown, Ad is = x√3, it follows, that the sum of all the perpendiculars, drawn from any point in an equilateral triangle to each of its sides, is equal to the whole perpendicular of the triangle. PROBLEM 6.-Through a given point P, in a given circle ACBD, to draw a chord CD, of a given length. But, by the property of the circle (Euc. 111. 35,) cp X PD = APX PB; whence Which equation, being resolved in the usual way, gives x = 1⁄2 a ± √ (1 a2 — bc); Where x has two values, both of which are positive. PROBLEM 7.-Through a given point P, without a giver circle ABDC, to draw a right line so that the part CD, intercepted by the circumference, shall be of a given length. B D Draw PAB through the centre o; and put CD=α, RÀ b, PB = c, and PC = x; then will PD = x + a. But, by the property of the circle, (Euc. III, 36, cor.,) PCX PD PA X PB; whence x (x -|- a) = bc, or x2 + ax = bc. Which equation being resolved, as in the former problem, gives XC 1⁄2 a ± √ (1 a2 + bc); Where one value of x is positive and the other negative.* PROBLEM 8.-The base of BC, of any plane triangle ABC, the sum of the sides AB, AC, and the line AD, drawn from the vertex to the middle of the base, being given, to determine the triangle. Put BD or DC= a, AD= b, AB + ACs, and AB = x; then will AC = S x. 2 But, by my Geometry, B. 11, Prop. 19, AB+ AC2 = 2BD +2AD2; whence x2 + (s —x)2 = 2a2 + 2b3, or SI · a2 + b2 — — s2. Which last equation, being resolved as in the former instances, gives #t 1⁄2 s ± s√ (a2 + b2 for the values of the two sides AB and taking the sign + for one of them, and 1 s2), 4 AC of the triangle; for the other, and observing that a2 + b2 must be greater than 1s2. PROBLEM 9.-The two sides AB, AC, and the line AD, bisecting the vertical angle of any plane triangle ABC, being given, to find the base BC. B The two last problems, with a few slight alterations, may be readily employed for finding the roots of quadratic equations by construction; but this, as well as the methods frequently given for constructing cubic and some of the higher order of equations, is a matter of little importance in the present state of mathematical science; analysis, in these cases, being generally thought a more commodions instrument than geometry. Put AB α, AC = b, ADC, and BC VI., 3, we shall have c, and BC= x; then, by Euc. AB (a): Ac (b):: BD: DC. And consequently, by the compositions of ratios, (Euc. v, 18,) But, by Euc. VI., 13, BD X DC + AD3 = AB X AC; where Which is the value of the base BC, as required. PROBLEM 10.-Having given the lengths of two lines AD, BE, drawn from the acute angles of a right-angled triangle ABC, to the middle of the opposite sides, it is required to determine the triangle. 2 Put AD α, BE = α, BE=b, b, CD or ‡CB = x, and CE or à ca = y; then, since (Euc. 1, 47) CD2 + CA2, = AD2, and CE2 — CB3 BE2, we shall have x2+4y2 Whence, taking the second of these equations from four times the first, there will arise And, in like manner, taking the first of the same equations from four times the second, there will arise |