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Geometry Difficulty 6.2 National olympiad Prove it Saudi Arabia

Let $ABC$ be an acute-angled triangle inscribed in the circle $(O)$ , $H$ the foot of the altitude of $ABC$ at $A$ and $P$ a point inside $ABC$ lying on the bisector of $\angle BAC$ . The circle of diameter $AP$ cuts $(O)$ again at $G$ . Let $L$ be the projection of $P$ on $AH$ . Prove that if $GL$ bisects $HP$ then $P$ is the incenter of the triangle $ABC$ .

Solution

Let $AP$ intersect $BC$ at $F$ and $(O)$ again at $D$ . Because $LP$ and $BC$ are parallel, we have
$\begin{aligned} \angle LGA &= \angle LPD = \angle BFD = \angle BAD + \angle CBA \\ &= \angle DAC + \angle CBA = \angle DBA = \angle DGA \end{aligned}$
This implies that $G$ , $L$ , $D$ are collinear.

Let $GL$ cut $BC$ at $E$ . Because $LE$ bisects $PH$ and $\angle EHL = \angle HLP = 90^\circ$ , quadrilateral $PLHE$ is a rectangle. We deduce, using Thales's theorem, that
$\frac{DP}{DA} = \frac{PE}{AL} = \frac{LH}{AL} = \frac{PF}{AP}$
Therefore, $\frac{DP}{DA} = 1 - \frac{AP}{DA} = 1 - \frac{PF}{DP} = \frac{DF}{DP}$ . We deduce that $DP^2 = DF \cdot DA$ .

On the other hand, $\angle DBF = \angle DAC = \angle BAF$ . This means that line $BD$ is tangent to the circumcircle of triangle $ABF$ and therefore $DB^2 = DF \cdot DA = DP^2$ . But $D$ is the circumcenter of triangle $BCI$ , where $I$ is the incenter of $ABC$ . We deduce that $P$ is the incenter of triangle $ABC$ .

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