Bump this beauty

Since $\overline{AQ} \parallel \overline{DJ}$, it suffices to prove that $\overline{XY}$ bisects $\angle APJ$. Let $\omega_X$ be the circle centered at $X$ tangent to $\overline{AB}$, $\overline{BC}$, and $\overline{DJ}$, and let $\omega_Y$ be the circle centered at $Y$ tangent to $\overline{AC}$, $\overline{BC}$, and $\overline{DJ}$. The following claim finishes. Claim: $\overline{AP}$ is tangent to $\omega_X$ and $\omega_Y$. (This is a special case of the main claim to ISL 2009 G8.) Proof: Let the tangent to $\omega_X$ passing through $A$ other than $\overline{AB}$ intersect $\overline{DJ}$ at $P'$. By (excircle) Pitot, $AC-CD=AB-BD=AP'-P'D$, so by the converse of Pitot, $ACDP$ is tangential, which means $\overline{AP'}$ is tangent to $\omega_Y$. Thus, $P'$ lies on $\overline{XY}$, so $P'=P$. $\square$

The problem is equivalent to show that the intersection of $AP$ and $BC$ lies on $(DXY)$ (or actually just on the perpendicular bisector of $XY$ is sufficient). Is there any solution based on this approach, which I find much more motivated than the above solution (it's beautiful, but still hard to think of)?

This problem was proposed by Burii.

does anyone have a solution based on the approach of a_507_bc ?

Oops...seems I solved the incircle version of the problem...although they're not different at all. Rename $J$ to be $I$ and the rest are defined similarly , with incircle though .We shall prove that the intersection of $AP$ and $BC$ lies on $(DXY)$ which finishes the problem after an easy angle chasing. Lemma(Serbia 2018). Let $Z,T$ be on $BI,CI$ st $\angle ZDT = 90$, Then $\angle ZAT = \angle BAI$. Proof. Trivial by moving points. Now , consider a point $Q$ on $ID$ st $ABDP$ is tangential with incircle $C(X,dist(X,BC))$. Then $AQ-QD = AB-BD = AC - CD$ Which means $ACDQ$ is tangential with incircle $C(Y,dist(Y,BC))$. So since $Q$ is the intersection of the common tangents of $C(X,dist(X,BC))$ and $C(Y,dist(Y,BC))$ , It would lie on $XY$ so $Q \equiv P$ Now , Let $R = AP \cap BC$. By the lemma , since $\angle XAY = \angle BAI$ we get that $\angle XDY =90$. But $X,Y$ are the incenters of $\triangle ABR, \triangle ACR$ respectively. So $\angle XRY =90$ and $R \in (XDY)$ so we're done.

It's also easy to coordbash this: Set $D=(0,0), B=(-b,0), C=(c,0)$ and $J=(0,-1)$ with $b,c>0$. Then we have $\ell_{BJ}: y=\frac{-1}{b}x-1$ and $\ell_{CJ}: y=\frac{1}{c}x-1$ and thus $X=\left(\frac{-b}{b+1},\frac{-b}{b+1}\right)$ and $Y=\left(\frac{c}{c+1},\frac{-c}{c+1}\right)$. Thus we have $\ell_{XY}: y=\frac{b-c}{2bc+b+c}x - \frac{2bc}{2bc+b+c}$. Also by tangent double angle we have $\ell_{AB}: y=\frac{2b}{1-b^2}(x+b)$ and $\ell_{AC}: y=\frac{2c}{c^2-1}(x-c)$, thus $A=\left(\frac{b-c}{bc-1},\frac{-2bc}{bc-1}\right)$. We need to show the reflection of $A$ over $XY$ lies on $JD$, or the foot from $A$ to $XY$ has $x$-coordinate $\frac{b-c}{2(bc-1)}$. The line from $A$ perpendicular to $XY$ has equation: $y=\frac{2bc+b+c}{c-b}x+\frac{b+c}{bc-1}$, intersecting this with $\ell_{XY}$, we have the foot from $A$ to $XY$ is $\left(\frac{b-c}{2(bc-1)},\frac{b+c-2bc}{2(bc-1)}\right)$, so we're done.

Let the $A$-excircle be tangent to sides $AB$ and $AC$ at points $E$ and $F$ respectively, furthermore let us introduce the point $Z$ on the ray $DJ$ with $DZ=AE=AF.$ $AQ\parallel{DJ},$ so we have to prove that the internal bisector of $\angle{APD}$ (which is parallel to the bisector of $\angle{QAP}$) is perpendicular to $XY,$ which is equivivalent to proving that $XY$ is the external bisector of $\angle{APD}.$ $D$ and $E$ are symmetric w.r.t. $BJ,$ so we have $\angle{ZDX}=\angle{XDB}=\angle{BEX}=\angle{AEX}=45°.$ We also now that $XD=XE$ (by symmetry), and $DZ=AE$ (by definition), hence $ZDX\triangle\simeq{AEX\triangle},$ from which we obtain $AX=ZX.$ Similarly $AY=ZY,$ hence $Z$ is the reflection of $A$ w.r.t. line $XY.$ This shows that lines $AP$ and $DPZ$ are symmetric w.r.t. $XY,$ hence $XY$ is one of their bisectors. Since $A$ and $D$ lie on the same side of $XY,$ it immediatly follows that it is the external bisector of $\angle{APD},$ yielding the desired result.