Problem - 3140
As shown, $\angle{ACB} = 90^\circ$, $AD=DB$, $DE=DC$, $EM\perp AB$, and $EN\perp CD$. Prove $$MN\cdot AB = AC\cdot CB$$
Let $\angle{ADE}=\alpha$ and $\angle{EDC}=\beta$.
Because $\angle{EMD}=\angle{END}=90^\circ$, points $EMDN$ are concyclic. By Ptolemy theorem: $$DE\cdot MD=ME\cdot DN + MD \cdot NE$$
\indent Let $R$ be the circumradius of $\triangle{ABC}$. Then $AB=2R, DE=DC=R$. It follows that
\begin{align*}
AB\cdot MN &= 2\cdot R \cdot MN \\
&= 2\cdot (DE \cdot MN)\\
&= 2\cdot (ME\cdot DN + MD \cdot NE)\\
&= 2\cdot (R\sin\alpha \cdot R\sin\beta + R\cdot\cos\alpha \cdot R\sin\beta)\\
&= 2R^2\sin(\alpha+\beta)\\
&= 2S_{\triangle{ABC}}\\
&=AC\cdot BC
\end{align*}