Let $\ell$ be the common $B$-midline of triangles $BCD$ (the point $B$ and the line $CD$ stay unchanged), and let $m$ be the common angle bisector of $\angle BCD$. Let $Z=\ell\cap m$. By Iran Lemma, $Z\in XY$ and so we are done.

Let line $\ell$ be angle bisector of $\angle ACB$ and $Z$ foot of perpendicular from $A$ to $\ell$. Then by Iran Lemma $Z$ is fixed point that all of $XY$ passes through it.

Yeah, similar to @above. The $B$-midline in $BCD$ is fixed (it's the midline in $ABC$) and the $\angle C$ bisector is also fixed, so by Iran Lemma, $XY$ passes through their intersection.

Let $I$ be the incenter of $\triangle ABC$. Let $P,R,Q$ be the incircle touchpoints with $BC,AB,AC$, respectively. Let $K=RQ\cap CI\cap (BRIP)$. Let $L$ be the foot from $D$ to $CK$. Let $S$ be the incenter of $\triangle DBC$. Let $Z$ be the foot from $S$ to $BC$. We claim that the fixed point is $K$. Firstly note that $L$ lies on $XZ$ by easy angle chase. By La Hire, it is sufficient to show that $D$ lies on the polar of $K$, which is equivalent showing that $SX$ is tangent to $(KXL)$. As $BKXSZ$ is cyclic due to right angles, we get that $$\measuredangle LKX=\measuredangle SKX=\measuredangle SZX=\measuredangle LXS,$$we are done. [asy][asy]import olympiad; size(10cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta; pair O,A,B,C,I,X,Y,D,P,Q,R,S,K,Z,L; O=(0,0);A=dir(140);B=dir(210);C=dir(330);path w=incircle(A,B,C);I=incenter(A,B,C);P=foot(I,B,C);Q=foot(I,A,C);R=foot(I,A,B); D=0.3A+0.7C;S=incenter(D,B,C);X=foot(S,B,D);Y=foot(S,C,D);K=extension(C,I,R,Q);Z=foot(S,B,C);L=foot(D,K,C); draw(A--B--C--cycle,deep);draw(w,heavyblue);draw(incircle(D,B,C),heavyblue);draw(D--B,deep);draw(K--C,deep);draw(R--Q,deep);draw(circumcircle(R,I,B),heavyblue); draw(circumcircle(B,K,X),heavyblue+dashed);draw(X--Z,deep+dashed);draw(L--D,deep);draw(X--S--Z,deep); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$I$",I,dir(I)); dot("$P$",P,dir(P)); dot("$Q$",Q,dir(Q)); dot("$R$",R,dir(R)); dot("$D$",D,dir(D)); dot("$S$",S,dir(S)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$K$",K,dir(K)); dot("$Z$",Z,dir(Z)); dot("$L$",L,dir(180)); [/asy][/asy]

I think this issue is much easier than the G3.

The bisector of $ \measuredangle ACB $ and the point of intersection of XY never change.

Hello sir would you please stop double and triple posting (use the edit tool) And please stop calling everything easy because if someone new fails in solving the problem it might hurt.