The trick is to note that the point of concurrence is actually a point of intersection of circumcircle of $BCP$ and line $AC$.

Trivial by complex numbers, let $A=a^2,B = b^2,C=c^2,P = \frac{2b^2c^2}{b^2 + c^2}$. Now consider both arc midpoints of $BA$, being $ab$, $-ab$. We then calculate the intersection of the line formed by the arc midpoints (which is just the perpendicular bisector), and $AC$. We can compute this as $\frac{b^2(a^2 + c^2)}{c^2 + b^2}$. The condition that this intersection point and $AB$ are parallel is equivalent to $\frac{a^2 - b^2}{\frac{b^2(a^2 + c^2)}{c^2 + b^2} - \frac{2b^2c^2}{b^2 + c^2}}$ being self conjugating which reduces to $\frac{(b^2 + c^2)(a^2-b^2)}{b^2a^2 - b^2c^2}$ self conjugating which is true.

Reflect $C$ over the perpendicular bisector of $AB$ to get $D$. Let $Q$, $R$ and $S$ be the intersections of the tangents to $\omega$ through $C$ and $D$, $D$ and $A$ as well as $A$ and $B$. Notice that the perpendicular bisector of $AB$ is just the line $QS$ and the line $PR$ is indeed parallel to $AB$ due to symmetry. Then, the result follows easily from the fact that in a quadrilateral with an incircle, the two diagonals and the connections of opposite points of tangency concur (for example, by Brianchon).

Use barycentric coordinates with respect to $ABC$, then $P=(-a^2:b^2:c^2)$, so the equation of the line through $P$ parallel to $AB$ is $c^2x+c^2y+(a^2-b^2)z=0$. The equation of the perpendicular bisector of $AB$ is $(a^2-b^2)z+c^2(x-y)=0$ and the equation of line $CA$ is $y=0$. From$$\begin{vmatrix}c^2&c^2&a^2-b^2\\c^2&-c^2&a^2-b^2\\0&1&0\end{vmatrix}=0$$we get that the three lines concur, as desired.

Let the perpendicular bisector of $\overline{AB}$ intersect $\overline{AC}$ at $K$, so $\triangle ABK$ is $K$-isosceles, and $\measuredangle PBC=\measuredangle BAK$, hence $\triangle ABK \sim \triangle CBP$, and therefore $\measuredangle BPC=\measuredangle BKA=\measuredangle BKC$, so $BCPK$ is cyclic. Thus, $\measuredangle PKC=\measuredangle BPC=\measuredangle BAC$, and thus $\overline{PK} \parallel \overline{AB}$, implying the desired concurrence. $\blacksquare$

Let $\ell$ be the perpendicular bisector of $AB$. Then, \[ -1 = (A, AP \,\cap\, \omega; B, C)_\omega \stackrel A= (AA \,\cap\, BB, AP \,\cap\, \ell; (A+B)/2, \ell \,\cap\, AC)_\ell \stackrel P= (B, A; (A+B)/2, P(\ell \,\cap\, AC) \,\cap\, AB)_{AB}.\]$\blacksquare$

Parallel to AB through P line intersect AC at X.We want to prove XA=XB.We note that <ACB=α and <BAC=β.Then we get <PBC=<PCB=β,<ABC=180-α-β.PX//AB then <BPX=α from angle chasing.<XCB=<XPB then XCPB is cyclic. From angle chasing <CPX=<CBX=180-α-2β.So we get <XΒA=β and XA=XB.We are done.

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