Product Of Opposite Sides In A Quadrilateral

The General Ptolemy states that if $\angle A+\angle C= \theta$ , then $(AC \cdot BD)^2=(AB \cdot CD)^2+(AD \cdot BC)^2-2 \cdot AB \cdot BC \cdot CD \cdot DA \cdot \cos \theta$ . Therefore, we have $(AC \cdot BD)^2=12^2+14^2-2 \cdot 12 \cdot 14 \cdot -\frac{2}{3}=\boxed{564}$

I took a parallelogram with sides $\sqrt14$ and $\sqrt12$ and one of the angles to be $\theta$ such that $\cos2\theta=-2/3$

Guide: Use the Ptolemaeus's theory. Draw the point E in the quadrilateral such that triangle ABE is similar to triangle DBC Then, we can prove that AB.CD + AC.BD = BD(EA + EC) = 26 and angle AEC = angle(A + C) Next, we use law of cosine in triangle AEC and the similarity ratio between (ABD, FBC) and (BCD,BFA) to find out the value