Saudi Arabia Mathematical Competitions

Let $K$, $L$, $M$ and $N$ be the feet of perpendiculars dropped from point $E$ to lines $AB$, $BC$, $CD$ and $DA$, respectively.

Points $K$ and $N$ lie on the circle with diameter $AE$, hence $\widehat{EKN}\equiv \widehat{EAN}$. Similarly, $\widehat{ELM}\equiv \widehat{ECM}\equiv \widehat{EAN}$, hence $$\begin{array}{c}\widehat{EKN}\equiv \widehat{ELN}.\end{array}\text{\hspace{1em}(1)}$$ In the same way we obtain $$\begin{array}{c}\widehat{ENK}\equiv \widehat{EML}.\end{array}\text{\hspace{1em}(2)}$$ From (1) and (2) it follows that $△EKN\sim △ELM$, therefore $$\frac{EK}{EN}=\frac{EL}{EM},\text{ hence }EN=\frac{EK\cdot EM}{EL}=\frac{ac}{b}.$$