Show that $3$ lines must concur

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 $$\angle BAC$$.

• $T$ doesn't need to lie on the angle bisector of $A$. If it does, then the centre of $\Omega$ would also lie on this angle bisector but there is absolutely no reason why it has to, unless $AB=AC$. Sep 30 '21 at 3:12
• @user10354138 I wasn't exactly sure of it, but I thought I had proved it, I guess I was wrong. Just curious, could I see an example? Sep 30 '21 at 3:26
• For every point O on the perpendicular bisector of BC that is sufficiently far away from BC in the opposite direction of A, the circle centred O through BC has a minor arc lying inside triangle ABC and hence you can construct a circle tangent to all three circlelines AB, AC, minor arc BC inside this shape. Sep 30 '21 at 3:41
• See this picture for example. Sep 30 '21 at 5:28 Let $$ω, Ω$$ have centers $$N, O$$ and radii $$r, R$$, and let $$D = BC ∩ MT$$. By the law of cosines on $$\triangle BON$$ and $$\triangle BOT$$, we have

$$\begin{multline*} BP^2 = BN^2 - r^2 = (R^2 + (R + r)^2 - 2R(R + r) \cos ∠BON) - r^2 \\ = \frac{R + r}{R}(R^2 + R^2 - 2R^2 \cos ∠BON) = \frac{R + r}{R} BT^2, \end{multline*}$$

and similarly $$CQ^2 = \frac{R + r}{R}CT^2$$. Since $$∠BTD$$ and $$∠MTC$$ subtend equal arcs of $$Ω$$, $$TD$$ is the external bisector of $$∠BTC$$, so $$\frac{BD}{CD} = \frac{BT}{CT} = \frac{BP}{CQ}$$. Therefore, $$\frac{AP}{PB} · \frac{BD}{DC} · \frac{CQ}{QA} = -1$$, which makes $$P, Q, D$$ collinear by Menelaus’s theorem.

• I almost completely understand what you are saying, I just don't understand the part where you go from an expression with a cosine in it to an expression without it ($\cos \angle BON$ to just $\angle BON$). Oct 4 '21 at 8:45
• @TymonMieszkowski That was a typo, fixed. Oct 4 '21 at 9:15
• @AndersKaseorg Can you please explain briefly why $TD$ is the external bisector of $\angle BTC$?
– YNK
Oct 4 '21 at 11:59
• @YNK $∠BTD$ subtends arc $BM$ of $Ω$. $∠CTM$ subtends arc $CM$ of $Ω$. $M$ is the midpoint of arc $BTC$, so arcs $BM$ and $CM$ are equal, so $∠BTD = ∠CTM$ by the inscribed angle theorem. If you’re confused by the unusual configuration of $∠BTD$, you can derive the same result from the more familiar configuration where the supplementary angle $∠BTM$ subtends everything except arc $BM$. If you’re still stuck here, review the inscribed angle theorem. Oct 5 '21 at 9:13
• @TymonMieszkowski It’s not a typo; the signed lengths are important to distinguish the very different configurations of Menelaus’s theorem and Ceva’s theorem. Other than that, this is just simple fraction manipulation combined with $AP = QA$. Oct 6 '21 at 2:27