Triangle $ABC$ has a right angle at $B$. Legs $\overline{AB}$ and $\overline{CB}$ are extended past point $B$ to points $D$ and $E$, respect

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 \(EB \cdot BD = AB \cdot BC\).