Nice geometry question to prove tangency Let $\triangle ABC$ be a scalene triangle with circumcircle $\omega$. The tangents to $\omega$ at $B$ and $C$ meet at a point $P$. Let $AP$ meet $BC$ at a point $K$ and let $M$ be the midpoint of $BC$. Let $X$ be a point on $AC$ such that $AB$ is tangent to the circumcircle of $\triangle BXC$. Let $BX$ meet $AM$ at a point $T$. Let the line through $C$ and parallel to $AM$ meet $TK$ at a point $V$.

Prove that $AV$ is tangent to $\omega$.

I tried the question with $\sqrt{bc}$ inversion with reflection over angle bisector. This approach does not seem to help much. Please provide hint?

First, $BX$ is parallel to the tangent to $\omega$ at $A$. So if $L=BX\cap AK$, we have $$-1=A(B,C;AK\cap\omega,A)=(B,X;L,\infty_{BX}).$$ So $L$ is the midpoint of $BX$, hence $L$ lies on $MN$, where $N$ is the midpoint of $AB$. Then $$-1=T(M,L;N,TK\cap MN)=(A,B;N,TK\cap AB),$$ so $TK\parallel AB$. Reflect $A$ over $M$ to $D$, so $CD\parallel AB\parallel TK$. Then $$-1=C(A,D;M,\infty_{AM})=(CA\cap TK,\infty_{TK};K,V)=(AC,AB;AK,AV).$$ This implies the result, as $(AC,AB;AK,AA)=-1$.
