Let $\omega_B$ be the circle with center $B$ and radius $AB$ and $\omega_E$ the circle of center $E$ and raidus $AE$. So $K=AX\cap \omega_B$ and $L=\omega_E\cap AX$. Let $F\in\omega_B$, $G\in \omega_E$ such that $F-A-G$ are aligned and this line is parallel to $CD$. Also let $P,Q$ be the second intersections of $FC$ and $GD$ with $\omega_B$ and $\omega_E$ respectively. Let $Z=FC\cap GD$, and $\ell_Z$ the line trough $Z$ parallel to $CD$ and so also to $FG$, and let $XY\cap \ell_Z=W$. By Reim's theorem, $CXKP$ and $DXLQ$ are cyclic. By Reim's theorem, $YPZW$ is cyclic and so is $YQZW$, or in other words $PYQZW$ is a cyclic pentagon. Therefore $XY$ always passes through the second intersection of $\ell_Z$ and $ZPQ$.

Really good one. Here's a solution with a background on how I solved it. I did it with geogebra, and without it I think I wouldn't be able to solve it. We rephrase the problem on the following way: let $X,Y$ be two circles that touch in $Z,W$. Let $Q,R$ be two fixed points on the plane and $P$ a variable point in $QR$. Let $PZ$ touch $X$ in $A$ and $Y$ in $B$. Prove that if $(APQ)\cap (BPR)=K$, then $PK$ has a fixed point. Firstly, when I drawed the real figure on geogebra (the one in the statement), I realized that the fixed point was the intersection of this line and the line between the two circles, so $A,B$ on the original statement were just hiding the condition. Then, I rephrased the problem in this way and I started to try to figure out what was this fixed point, but I wasn't able to do so. But then I stopped trying it and two days later (today) I thought: what if I drew two points $P$ and try to prove that the lines concur? And this is what I did, and for my surprise lots of things appeared in my new drawing. Let $L,M$ be the intersections between $X$ and $(APQ)$ and $Y$ and $(BPR)$. First, we show that $L,M$ are fixed. For this, just take another point $P'$ on $QR$ and show that $LP'A'Q$ is cyclic, where $A'=ZP'$. For this, just notice that $180- \angle LA'P'=180-\angle LAZ=\angle LQP$. Then I tried to find any relation between $L,M$ and the drawing. Then, if you call the point which we want by $T=ZW\cap PK$ (we want it to be fixed), then if we connect $TM,TL$ with $(APQ),(BPR)$ respectively then this new intersection points are collinear with $P$. Then I realized this was true iff $KLWT$ was cyclic. Then I drawed it's circumcircle and the magic happened: $WLMKT$ is cyclic. We prove now that both $M,L$ lie both in $(WKT)$. Notice that if we show this, we show that $T$ lies on $ZW$ and on $(WML)$, whence it is fixed. But just notice $\angle MKT=180-\angle MKP=\angle MBP=180-\angle MBZ \implies \angle MBZ=\angle MKT$, but $\angle MBZ=\angle MWZ=\angle MWT$ and we're done. $\blacksquare$