[asy][asy] import olympiad; import geometry; size(13cm); pair A= dir(104); pair B=dir(230); pair C=dir(310); pair I=incenter(A, B, C); pair O=excenter(B, C, A); pair D=foot(A, B, C); pair E=foot(I, B, C); pair X=intersectionpoint(line(E, O), line(A, D)); pair F=foot(O, B, C); pair P=circumcenter(X, B, C); pair J1[]=intersectionpoints(line(X, I), circle(O*2-F, F)); pair J=J1[0]; dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, SE); filldraw(A--B--C--cycle, invisible, red); draw(circle(I*2-E, E), orange); draw(circle(O*2-F, F), orange); draw(A--D, red); draw(X--O, red); draw(A--(A+(B-A)*1.8), red); draw(A--(A+(C-A)*1.5), red); dot("$I$", I, NE); dot("$O$", O, S); dot("$D$", D, SW); dot("$E$", E, SW); dot("$X$", X, NW); dot("$F$",F, S); dot("$P$", P, NE); dot("$J$", J1[0], N); draw(X--J, magenta+dashed); draw(B--X--C, red); draw(circle(B, J, C), lightred+dashed); [/asy][/asy] Let $F$ be the touchpoint of the excircle at $BC$, and let $J$ be the point on the $A$-excircle so that $\odot(BJC)$ is tangent to the excircle. It suffices to prove that $PJ$ bisects $\angle BJC$, since this would imply that $P$ is the midpoint of arc $BC$ of $\odot(BJC)$. We start with recalling ISL 2002 G7 from which we know that $OX$ bisects $\angle BXC$. Since $X$ lies on the incircle, we have $\angle DXE=\angle XEI=\angle EXI$; thus $XI$ and $XD$ are isogonal WRT $\angle BXC$. Hence $X, I, P$ are collinear. Note that it is well-known that $F$ lies on this line as well. Now it suffices to prove that $F, I, J$ are collinear, and that $IJ$ bisects $\angle BJC$. But this is the excenter version of ISL 2002 G7 (which can be proved analogously), so we're done.

It's known that $X$ coincides with the midpoints of $\overline{AD}$ and also lies on line $\overline{IF}$. [asy][asy] size(13cm); pen zzttqq = rgb(0.6,0.2,0); pen fuqqzz = rgb(0.95686,0,0.6); pen qqwuqq = rgb(0,0.39215,0); pen cqcqcq = rgb(0.75294,0.75294,0.75294); draw((-2.43454,2.03316)--(-2.6,0)--(-0.2,0)--cycle, linewidth(1) + zzttqq); draw((-2.43454,2.03316)--(-2.6,0), linewidth(1) + zzttqq); draw((-2.6,0)--(-0.2,0), linewidth(1) + zzttqq); draw((-0.2,0)--(-2.43454,2.03316), linewidth(1) + zzttqq); draw(circle((-1.4,0.92565), 1.51553), linewidth(1)); draw(circle((-1.89059,0.65401), 0.65401), linewidth(1) + fuqqzz); draw((-2.43454,2.03316)--(-2.43454,0), linewidth(1)); draw(circle((-0.90940,-1.83376), 1.83376), linewidth(1) + qqwuqq); draw((-2.43454,0)--(-1.4,-0.58987), linewidth(1)); draw((-2.43454,2.03316)--(-1.4,-0.58987), linewidth(1)); draw(circle((-1.4,0.32644), 1.24360), linewidth(1) + qqwuqq); draw((-2.6,0)--(-2.73712,-1.68502), linewidth(1)); draw((-2.43454,1.01658)--(-0.90940,-1.83376), linewidth(1)); draw((-0.90940,-1.83376)--(-0.90940,0), linewidth(1)); draw((-1.4,-0.91688)--(-0.90940,0), linewidth(1)); dot("$A$", (-2.43454,2.03316), dir((1.347, 3.366))); dot("$B$", (-2.6,0), dir(225)); dot("$C$", (-0.2,0), dir(-45)); dot("$I$", (-1.89059,0.65401), dir((1.479, 2.608))); dot("$D$", (-2.43454,0), dir((1.347, 2.711))); dot("$E$", (-1.89059,0), dir((1.479, 2.711))); dot("$M$", (-1.4,-0.58987), dir((1.224, 2.797))); dot("$O$", (-0.90940,-1.83376), dir((1.306, 2.649))); dot("$X$", (-2.43454,1.01658), dir(135)); dot("$P$", (-1.4,0.32644), dir((1.224, 2.716))); dot("$F$", (-0.90940,0), dir((1.306, 2.711))); dot("$N$", (-1.4,-0.91688), dir(225)); [/asy][/asy] Claim: $(BXC)$ is tangent to the incircle and passes through the midpoint $N$ of $\overline{EO}$. Proof. Follows by 2002 G7. $\blacksquare$ Hence by homothety at $X$ the line $\overline{XIF}$ passes thru $P$. Claim: $(BXNC)$ and the $A$-excircle are orthogonal. Proof. It suffices to show $BXCN$ is fixed under inversion around the $A$-excircle. To this end, we prove that \[ ON \cdot OX = OF^2. \]Indeed, this follows from the similar isosceles triangles \[ \triangle ONF \sim \triangle OFX \sim \triangle EIX. \]$\blacksquare$ Hence it follows that inversion centered at $P$ with radius $PB = PC$ will fix the $A$-excircle. Since $BC$ is tangent to the $A$-excircle at $F$, the inverse image of $F$ is the desired tangency point.