# An Identity in (Cyclic) Quadrilaterals

### Problem

Let $ABCD$ be a cyclic quadrilateral; $E$ the intersection of the diagonals.

Then

$\displaystyle\frac{AE}{CE}=\frac{AB}{BC}\cdot\frac{AD}{CD}.$

### Acknowledgment

The problem has been posted by Dao Thanh Oai (Vietnam) at the CutTheKnotMath facebook page.