Given $\triangle ABC$, whose all sides have different length. Point $P$ is chosen on altitude $AD$. Lines $BP$ and $CP$ intersect lines $AC, AB$ respectively and point $X, Y$.It is given that $AX=AY$. Prove that there is circle, whose centre lies on $BC$ and is tangent to sides $AC$ and $AB$ at points $X,Y$.
We begin with this following crucial claim.
Claim: Quadrilateral $AXDY$ is cyclic.
Proof: Let $XY \cap BC=Z$. We have that:
$$ -1 = (Z,D;B,C) \overset{A}{=} (Z, AD \cap XY;X,Y) $$Now since $\angle ZDA = 90^{\circ}$ we must have that $\angle ADY = \angle ADX$. By sinus theorem in triangles $\triangle ADY$ and $\triangle ADX$ we get that $\sin \angle AYD = \sin \angle AXD$. It is impossible to have $\angle AYD = AXD$ since this will force to have that $AB = AC$, which is clearly impossible. Thus we get that $\angle AYD+\angle AXD =180^{\circ}$ as desired.
Now let $T$ be a point on side $BC$ such that $TY \perp AB$. This forces to $T$ lie on $\odot(AYDX)$. Now we are done because we have that $TX \perp AC$ and $\angle AYX = \angle AXY \implies \angle TYX = \angle TXY \implies TX=TY$, which proves existence of circle.