Here’s a diagram for anyone that wants to solve this:
(I’m kind of bad at proving concurrency so oh well)

These are so cool - would you be alright with me using a diagram on the login page of the website? It’s pretty dry right now haha

This ones cool

Here we go!

I have drawn in \overline{PZ} and \overline{NY} with their intersection at O.

Now O lies on the circumcircle of ABC because ZCNO is an isosceles trapezoid.

So to prove the concurrency, we can prove that \angle OMC + \angle XMC = 180.
We know that \angle XMC =90 as \overline{XM} is the perpendicular bisector of \overline{BC}.

Let’s angle chase to find \angle OMC!
Setting the angles of the triangle as 2a, 2b, 2c we get that \overline{PM} is perpendicular to the bisector of the C-excircle. (Easy to prove as the bisector of C in \triangle ABC is perpendicular to the bisector of the C-excircle.

Now we can look at \angle ZOY which is equal to 2a+b+c (looking at the arcs it intersects).
From here we can also get that \angle MNQ = 2a+b+c because of our parallel lines.
Now since \angle CNQ = 90 = a+b+c, \angle CNM = a and thus \angle MNO = b+c.

\angle MNO = b+c implies that \angle BYO = b+c.
\angle BYO intersects an arc of angle 2c+q. (q=\angle ACO)

Isolating q we get q=b-c.

Adding 2c we get that \angle MCO = b+c.
That’s equal to \angle MNO!
This means that OMCN is a cyclic quadrilateral and thus \angle OMC = 180 - \angle CNO = 180-90 =90 and we can conclude that \overline{XMO} is a straight line.


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:exploding_head: :exploding_head: :exploding_head: :exploding_head: :exploding_head: Honestly this is incredible, well done!!! :exploding_head: :exploding_head: :exploding_head: :exploding_head: :exploding_head: