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?
 A: 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 $(AC,AB;AK,AA)=-1$.
A: Here you have an approach employing barycentric coordinates. For more information, have a look at this awesome handout by Evan Chen. Following his advice, I will use , when referring to normalized coordinates, and : for homogeneous coordinates.
Let $\triangle ABC$ be the reference triangle, i.e. $A=(1,0,0); B=(0,1,0);C=(0,0,1)$ with side lengths $a=BC, b=CA, c=AB$. Observe that $M=(0:1:1).$ Furthermore, notice that $(\triangle BXC)$ being tangent to $AB$ is equivalent to $$\angle XBA=\angle ACB\implies \triangle ABX\sim \triangle ACB\implies \frac{AX}{AC}=\frac{c^2}{b^2} $$ Therefore $$\overrightarrow{AX}=\frac{c^2}{b^2}\cdot \overrightarrow{AC}=\frac{c^2}{b^2}\cdot (-1, 0, 1)=\left(-\frac{c^2}{b^2}, 0, \frac{c^2}{b^2}\right)\implies X=\left(1-\frac{c^2}{b^2},0, \frac{c^2}{b^2}\right)\equiv \left(b^2-c^2:0:c^2\right)$$
Thus, we infer that $AM: z-y=0$ and $BX: c^2\cdot x-\left(b^2-c^2\right)\cdot z=0$. It follows that $T=\left(b^2-c^2:c^2:c^2\right)$.
On the other hand, looking at the definition of $K$, it is well-known that $AK$ is but the $A$-symmedian. Hence, $$AK:b^2z-c^2y=0, \; BC:x=0\implies AK\cap BC=K=\left(0:b^2:c^2\right)$$ It follows that \begin{align*}TK&: (b^2-c^2)b^2\cdot z+c^4\cdot x-c^2b^2\cdot x-(b^2-c^2)c^2\cdot y=0\\\iff TK&: (b^2-c^2)\cdot (b^2z-c^2y)+c^2x\cdot (c^2-b^2)=0\\\iff TK&: b^2z-c^2y-c^2x=0\end{align*} In order to determine the equation of the line $l$ through $C$ parallel to $AM$, I will use the method described here (see page $76$). This yields $$l: x+2y=0$$ Thus $l\cap TK=V=(-2b^2:b^2:-c^2)$. Besides, the equation of the tangent $t$ to $\omega$ at $A$ is given by $b^2z+c^2y=0$ (cf. Chen's handout, corollary $15$). We are almost there, since $$V\in t\iff b^2\cdot(-c^2)+c^2\cdot b^2=0 $$ Which is clearly true.
