# 2019 USAJMO Problems/Problem 3

## Problem

$(*)$ Let $ABCD$ be a cyclic quadrilateral satisfying $AD^2+BC^2=AB^2$. The diagonals of $ABCD$ intersect at $E$. Let $P$ be a point on side $\overline{AB}$ satisfying $\angle APD=\angle BPC$. Show that line $PE$ bisects $\overline{CD}$.

## Solution

Let $PE \cap DC = M$. Also, let $N$ be the midpoint of $AB$.

Note that only one point $P$ satisfies the given angle condition. With this in mind, construct $P'$ with the following properties:

$AP' \cdot AB = AD^2 \quad \text{and} \quad BP' \cdot AB = CD^2$

Claim:$P = P'$

Proof:

The conditions imply the similarities $ADP \sim ABD$ and $BCP \sim BAC$ whence $\measuredangle APD = \measuredangle BDA = \measuredangle BCA = \measuredangle CPB$ as desired. $\square$

Claim: $PE$ is a symmedian in $AEB$

Proof:

We have

$$AP \cdot AB = AD^2 \iff AB^2 \cdot AP = AD^2 \cdot AB$$ $$\iff \left( \frac{AB}{AD} \right)^2 = \frac{AB}{AP}$$ $$\iff \left( \frac{AB}{AD} \right)^2 - 1 = \frac{AB}{AP} - 1$$ $$\iff \frac{AB^2 - AD^2}{AD^2} = \frac{BP}{AP}$$ $$\iff \left(\frac{BC}{AD} \right)^2 = \left(\frac{BE}{AE} \right)^2 = \frac{BP}{AP}$$

as desired. $\square$

Since $P$ is the isogonal conjugate of $N$, $\measuredangle PEA = \measuredangle MEC = \measuredangle BEN$. However $\measuredangle MEC = \measuredangle BEN$ implies that $M$ is the midpoint of $CD$ from similar triangles, so we are done. $\square$

~sriraamster