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 X. Then the triangle Aẞy is similar to ABC; so that COR. 1. Square on diameter of Brocard Circle = R2 - 3p2. COR. 2. Since K and 'K subtend angles equal to w in the Brocard Circle, .. QK = Q'K = SK sin w = R sin w √(1 − 3 tan2 w), 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. Then ABCD= = ΔΑΒD + Δ BCD = (AB + DC) (perpendicular distance between AB and CD), = (sum of parallel sides) × altitude. This may be written A B 1 (AB+CD) AD sin ADC or (AB+ CD) BC sin BCD. = B (1). (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) (sd)}......(2). 224. The area of any quadrilateral (in terms of its sides and diagonals). 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 + ď2 — c2 = 2 (AO. OB + BO. OC .. 16Q2 (2ef+b2 + d2 — a2 - c2) (2ef + a2 + c2 - b2 - d)... (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.) A Let A + C = 20, so that B+D=2(π-0). Then Now Q=A ABD + BCD = ad sin A+ bc sin C.......(1). BD2 = a2 + d2-2ad cos A = b2 + c2 - 2bc cos C. .. a2 - b2 + d2 - c2 = 2ad cos A - 2bc cos C. .. (a2 — b2 + d2 — c2)2 = 4a2d2 cos2 A - 8abcd cos A cos C + 4b2c2 cos2 C, 16Q2 = 4a2d2 sin2 A + 8abcd sin A sin C + 4b2c2 sin2 C. and .. 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) (8b) (sc) (8 d) - abcd cos3 0......(2). Q COR. 1. If is inscribable in a circle, 20, 0= and cos = 0. Hence Q = {(8-a) (s—b) (8 - c) (8 d)}. COR. 2. If a circle is inscribable in Q, 8-a=c, 8-b=d, s-c=a, 8. d=b. .. Q= √(abcd). sin 0. COR. 3. A quadrilateral of given sides has its maximum area when it is inscribable in a circle. |