The tangents of the circumcircle $\Omega$ of the triangle $ABC$ at points $B$ and $C$ intersect at point $P$. The perpendiculars drawn from point $P$ to lines $AB$ and $AC$ intersect at points$ D$ and $E$ respectively. Prove that the altitudes of the triangle $ADE$ intersect at the midpoint of the segment $BC$.
parmenides51 wrote:
The tangents of the circumcircle $\Omega$ of the triangle $ABC$ at points $B$ and $C$ intersect at point $P$. The perpendiculars drawn from point $P$ to lines $AB$ and $AC$ intersect at points$ D$ and $E$ respectively. Prove that the altitudes of the triangle $ADE$ intersect at the midpoint of the segment $BC$.
Let's draw altitude $DX$ of $ADE$ from $D$ to $AE$. Let $DX$ intersects $BC$ at point $T$, and $ET$ intersects $AB$ at $Y$. it suffices to prove that $EY \perp AB$. Let's say $\angle BAC = \alpha$. since $PB$ is tangent to the circle we get $PBT = PBC = \alpha$. Because of $AXD = 90^o$, we have $\angle CDA = 90^o-\alpha \Longrightarrow \angle PDT = PDX = \alpha$. Since $PBT = \alpha = PDT$ we have $PDBT$ is cyclic. So $\angle PTB = 90^o \Longrightarrow \angle PTC = 90^o \Longrightarrow \angle PTC+ \angle PEC = 180^o \Longrightarrow PTCE$ is cyclic. Since $PC$ is tangent we also have $\angle PCB = \alpha \Longrightarrow \angle PET = \alpha \Longrightarrow \angle AEY = 90^o-\alpha \Longrightarrow \boxed{EY \perp AB}$