Saudi Arabian Mathematical Competitions

Let $K$, $L$, $M$ and $N$ be the feet of perpendiculars dropped from point $E$ to lines $A B$, $B C$, $C D$ and $D A$, respectively.

Points $K$ and $N$ lie on the circle with diameter $A E$, hence $\widehat{E K N} \equiv \widehat{E A N}$. Similarly, $\widehat{E L M} \equiv \widehat{E C M} \equiv \widehat{E A N}$, hence
$$\begin{equation*} \widehat{E K N} \equiv \widehat{E L N} . \tag{1} \end{equation*}$$
In the same way we obtain
$$\begin{equation*} \widehat{E N K} \equiv \widehat{E M L} . \tag{2} \end{equation*}$$
From (1) and (2) it follows that $\triangle E K N \sim \triangle E L M$, therefore
$$\frac{E K}{E N}=\frac{E L}{E M}, \text{ hence } E N=\frac{E K \cdot E M}{E L}=\frac{a c}{b} .$$