Problem:
Let ABCD be a convex quadrilateral with AB=CD=10,BC=14, and AD=265. Assume that the diagonals of ABCD intersect at point P, and that the sum of the areas of △APB and △CPD equals the sum of the areas of △BPC and △APD. Find the area of quadrilateral ABCD.
Solution:
Let a,b,c, and d denote AP,BP,CP, and DP, respectively, and let θ=∠CPD. The statement about equal areas says that 21(ab+cd)sinθ=21(ad+bc)sin(π−θ), which implies (a−c)(d−b)=0. The cases where a=c and b=d are similar, so assume that a=c. By the Law of Cosines
a2+b2+2abcosθ=196 and a2+b2−2abcosθ=100
so a2+b2=148 and abcosθ=24. Also
a2+d2+2adcosθ=260 and a2+d2−2adcosθ=100,
so a2+d2=180 and adcosθ=40. Thus d2−b2=32 and bd=35, which yields d=52,b=32,a=130, and cos2θ=6516 and sin2θ=6549. The requested area is