I was recently posed with the following problem:
Let $\Omega$ be a circle passing through vertices $B$ and $C$ in triangle $\Delta ABC$ and let $\omega$ be a circle tangent to segments $AB$ and $AC$ at the points $P$ and $Q$ respectively, and externally tangent to $\Omega$ at point $T$. Let $M$ be the midpoint of arc $BTC$ of $\Omega$. Show that lines $BC$, $PQ$, and $MT$ concur.
Naturally, I first made a quick sketch of the problem:
I then quickly realised that in order for $\Omega$ to pass through both $B$ and $C$, its centre must lie on their perpendicular bisector. This also quickly leads to finding $M$, as it is simply the point where $\Omega$ intersects this perpendicular bisector.
I then managed to loosely show that $T$ has to be on the angular bisector of $\angle BAC$, i.e. the point where $\Omega$ intersects with this angular bisector. This also very simply shows that $P$ and $Q$ must be the same distance away from $A$, or in other words, that the triangle $\Delta PQA$ is isosceles.
With this information, I made a new, more accurate diagram:
I then started trying to prove the statement in the original problem. Since we know $M$ and $T$, and we know $BC$, we can now find the point where the three lines must concur. Then, since we know that $\Delta PQA$ is isosceles, my idea was that we could find the point of concurrence, and draw a line from it to $\Delta ABC$, such that it would make an isosceles triangle at the points where it intersects with the triangle, and check if those points are $P$ and $Q$ (the points of tangency with the circle). If they are, then we have proved it to be true, since $P$ and $Q$ make a line to the point of concurrence, otherwise, it is false.
This is where I got stuck, I don't know how I would go about finding this line from the point of concurrence to $\Delta ABC$, or how I would then go about showing that it intersects at $P$ and $Q$ specifically. I pondered this for a few days but couldn't think of anything, so I decided to ask it here. All help is greatly appreciated!
EDIT: it turns out a claim I made earlier was wrong, $T$ does not have to lie on the angular bisector as @user10354138 pointed out
EDIT: I also had another finding I forgot to include in my original question. Since $\omega$ is tangent to line $AB$ at the point $P$, it follows that the centre of $\omega$ must lie on the line perpendicular to $AB$ at point $P$, and similarly for $AC$ and point $P$, like so:
And since the lines $OP$ and $OQ$ (where $O$ is the centre of $\omega$) must be the same length, $O$ must lie on the angular bisector of $\angle BAC$.