Let $ABCD$ be a convex quadrilateral, and write $\alpha=\angle DAB$; $\beta=\angle ADB$; $\gamma=\angle ACB$; $\delta= \angle DBC$; and $\epsilon=\angle DBA$. Assuming that $\alpha<\pi/2$, $\beta+\gamma=\pi /2$, and $\delta+2\epsilon=\pi$, prove that \[(DB+BC)^2=AD^2+AC^2\] [Moderator edit: Also discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=30569 .]