Question 16
(CZS 5) A convex quadrilateral $ABCD$ with sides $AB=a$, $BC=b$, $CD=c$, $DA=d$ and angles $\alpha=\angle DAB$, $\beta=\angle ABC$, $\gamma=\angle BCD$, and $\delta=\angle CDA$ is given. Let $s=(a+b+c+d)/2$ and $P$ be the area of the quadrilateral. Prove that
$$P^2=(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\frac{\alpha+\gamma}{2}.$$
$$P^2=(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\frac{\alpha+\gamma}{2}.$$