COR. 1. This is another proof that the nine-points circle touches the incircle. COR. 2. If we write I, for I, r1 for r, and + for in the above we have the similar proposition for the escribed circle. 220. The distance between the Circumcentre and the Cosinecentre. Join AK. Through K draw By antiparallel to BC. Join SA cutting By at right-angles in λ. Then the triangle Aẞy is similar to ABC; so that COR. 1. Square on diameter of Brocard Circle = R2 – 3p2. COR. 2. Since OK and 'K subtend angles equal to w in the Brocard Circle, .. QK = Q'K = SK sin w = R sin ∞ √√(1 − 3 tan2 w), and NS='S = SK cos ∞ = R√(cos2 - 3 sin2 ∞), and 20′ = SK sin 2w = R sin 2∞ √(1 − 3 tan2 w). This last result was reached in Art. 215. 222. Area of a trapezium (i.e. a quadrilateral with one pair of opposite sides parallel). Let ABCD be a quadrilateral in which AB, CD are parallel. (AB+ CD) AD sin ADC or (AB+ CD) BC sin BCD. 223. Area of a quadrilateral inscribable in a circle. = = (a + d + b − c) (a + d − b + c) (b + c + a − d) (b + c − a + d), .. dividing by 16 and taking the square root Quadrilateral ABCD = √√{(8 − a) (8 — b) (s — c) (s — d)}................(2). 224. The area of any quadrilateral (in terms of its sides and diagonals). D C A Let Q area of ABCD. = B AB-a, BC=b, CD = c, DA = d, AC = e, BD =ƒ. Let AC, BD cut in O. Then Q=AOB+ BOC + COD + DOA b2 = BO2 + CO2-2BO. CO cos BOC, .. b2 — a2 + d2 — c2 = 2 (AO . OB + BO . OC = 2ef cos AOB. + CO. OD + DO. OA) cos AOB .. 2ef (1 + cos A OB) = 2ef + b2 + ď2 — a2 -- c2, 2ef (1 - cos AOB) = 2eƒ + a2 + c2 — b2 — ď2, .. 16Q2= (2ef+b2 + d2- a2 - c2) (2ef + a2 + c2-b2 - d3)... (2). Cor. 1. Since in a circle ef=ac + bd, (Euc. VI. D) therefore his equation reduces to the formula (2) of the preceding article. Cor. 2. If a circle is inscribable in the Quadrilateral, 225. The area of any quadrilateral: (in terms of its sides and the sum of its opposite angles.) .. (a2 — b2 + d2 — c2)2 = 4a2ď2 cos2 A - 8abcd cos A cos C + 4b2c2 cos2 C, and 16Q2 = 4a2d2 sin2 A + 8abcd sin A sin C + 4b2c2 sin2 C. .. 16Q2 + (a2 − b2 + ď2 — c2)2 = 4a2ð2 – 8abcd cos (A + C') + 4b2c2. .. 16Q2 = 4 (ad + bc)2 — (a2 — b2 + d2 - c2)2 - 16abcd cos2 0 = (2ad+2bc+a2-b2 + d2 - c2) (2ad+2bc-a2 + b2 - d2 + c2) .. Q2 = (8 − a) (8—b) (8 - c) (8 − d) - abcd cos2 0......(2). = COR. 1. If is inscribable in a circle, 20π, and cos 0 = 0. Hence Q√(sa) (s—b) (8 - c) (8 d)}. = COR. 2. If a circle is inscribable in Q, s—a = c, s-b=d, s− c = a, s :. Q= √(abcd). sin 0. d=b. COR. 3. A quadrilateral of given sides has its maximum area when it is inscribable in a circle. |