Question 103
Let $ABCD$ be a convex quadrilateral. Put $\angle DAB=\alpha$, $\angle ADB=\beta$, $\angle ACB=\gamma$, $\angle DBC=\delta$, $\angle DBA=\epsilon$. Given that $\alpha\lt 90^\circ$, $\beta+\gamma=90^\circ$, and $\delta+2\epsilon=180^\circ$, prove that
$$(DB+BC)^2=AD^2+AC^2.$$
$$(DB+BC)^2=AD^2+AC^2.$$