Triangle \(ABC\) has a right angle at \(B\). Legs \(\overline{AB}\) and \(\overline{CB}\) are extended past point \(B \) to points \(D\) and \(E\), respectively, such that \(\angle EAC = \angle ACD = 90^\circ\). Prove that \(EB \cdot BD = AB \cdot BC\).
Draw a picture!
