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Geometry Difficulty 6.3 National olympiad Prove it Saudi Arabia

Let $A B C$ be a triangle with orthocenter $H$ . Let $P$ be any point of the plane of the triangle. Let $\Omega$ be the circle with the diameter $A P$ . The circle $\Omega$ cuts $C A$ and $A B$ again at $E$ and $F$ , respectively. The line $P H$ cuts $\Omega$ again at $G$ . The tangent lines to $\Omega$ at $E, F$ intersect at $T$ . Let $M$ be the midpoint of $B C$ and $L$ be the point on $M G$ such that $A L$ and $M T$ are parallel. Prove that $L A$ and $L H$ are orthogonal.

Solution

Let $B Y, C Z$ be altitudes of $A B C$ . Points $Y, Z$ lie on the circle of diameter $A H$ . The line $H P$ cuts the circle $\Omega$ again at $G$ . Since $A P$ is a diameter in $\Omega$ , the lines $A G$ and $P H$ are perpendicular and therefore point $G$ lies on the circle of diameter $A H$ .

Because $A, E, G, F$ are concyclic and $A, G, Z, H, Y$ are concyclic, we have $\angle G E F=\angle G A F=\angle G A Z=\angle G Y Z$ and $\angle F G E=\angle Z A Y=\angle Z G Y$ . We deduce that triangles $G E F$ and $G Y Z$ are similar.
We know that $\angle M Z C=\angle Z C B=\angle Z A H$ . We deduce that $Z M$ is tangent to the circumcircle of triangle $G Z Y$ . Similarly, $M Y$ is tangent to the circumcircle of triangle $G Z Y$ . But $F T$ and $E T$ are tangent to the circumcircle of triangle $G F E$ . We deduce that quadrilaterals $G F T E$ and $G Z M Y$ are similar and therefore $\angle G Z A=\angle G Z F=\angle G M T=\angle G L A$ , since $L A$ is parallel to $M T$ . This means that point $L$ lies on the circle of diameter $A H$ and therefore $L A$ and $L H$ are orthogonal.

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