Let $D=BQ\cap AP$, $\angle BAC=\alpha$, $\angle ABC=\beta$, $\angle ACB=\gamma$, $\ell$ the bisector of $\angle BAC$, $J=\ell\cap BC$ Fact 1: $Y$ is the midpoint of $RD$ Proof: since $QX=XB$ and $AQ=AB$, we have $AX\perp BQ$. $\angle AQB=\angle ABQ=\frac{\alpha}{2}$ $\Rightarrow$ $BD\parallel \ell$. $\triangle PAJ$ is isosceles because $\angle PAJ=\angle PAB+\angle BAJ=\gamma+\frac{\alpha}{2}$ and $\angle AJP=\angle AJB=\pi-\angle BAJ-\angle ABJ=\pi-\beta-\frac{\alpha}{2}=\gamma+\frac{\alpha}{2}$. Then $AP=PJ$ It follows that $BDAJ$ is an isosceles trapezoid, with $AD=BJ$. Since $AP=PJ$, $AR=PB$, then $PR=BJ=AD$ which is equivalent to $RY=YD$ as $PY=AY$ $_{\blacksquare}$ Let $H$ be the midpoint of $QR$ and $G$ be the midpoint of $BR$. From fact 1 it follows that $H,Y,G$ are collinear and $HY\parallel BQ\parallel \ell$, so $HY\perp AX$ Fact 2: $AH\parallel BC$ Proof: $AQDJ$ is a parallelogram because $\angle DQA=\angle DBA=\angle DJA=\frac{\alpha}{2}$ (the last equality is due to the fact that $ADBJ$ is an isosceles trapezoid) Let $L=HY\cap BC$. $YL=\frac{1}{2}AJ$ as $PY=AY$. $HY=frac{1}{2}DQ=\frac{1}{2}AJ=YL$ Since $HY=YL$ and $PY=AY$, we have that $PHAL$ is a parallelogram. Therefore $AH\parallel BC$ $_{\blacksquare}$ Fact 3: $H$ and $Y$ are symmetric wrt $AX$ Proof: $AHLJ$ is a parallelogram since $AH\parallel BC$ and $HL\parallel AJ$. Since $PY=AY$, we have $PL=LJ$ Then $AH=LJ=AY$. Since $AH=AY$ and $HY\perp AX$, we get that $H$ and $Y$ are symmetrical wrt $AX$ $_{\blacksquare}$ As a result, $XY=HX=GB=\frac{1}{2}BR$ ($HX=GB$ because $H,X,G$ are the midpoints of $RQ,QB,BR$)

Let $S$ be the intersection of segments $BE$, $AP$. Since $AC>AB\implies \angle ABC>\angle ACB$ and $AX\perp BE$ we have $\angle XAB=\frac{\angle ABC+\angle ACB}{2}>\angle ACB=\angle SAB$ which proves $S$ belongs to the segment $BX$. $$AE=AB\implies \angle ABS=\frac{\angle BAC}{2}\implies \angle ASB=180^\circ-\frac{\angle BAC}{2}-\angle ACB=\frac{\angle BAC}{2}+\angle ABC=\angle SBC$$Therefore $AR=PB=PS$ and $Y$ is the midpoint of segment $RS$. Let $T$ be the midpoint of segment $BR$. We will prove $XY=BT$. Denote $G$ as intersection of lines $YT,BC$. Segment $YT$ is a midline in triangle $BSR$ so $GY\parallel BS$ and $BG=SY$. Because $\angle TGB=180^\circ-\angle GBS= \angle YSX$ we need only $SX=TG$ to get $\triangle TGB\equiv\triangle XSY$. Let's focus on this equality. Menelaos in $\triangle PGY$ with line $B,T,R$ gives $$1=\frac{PB}{BG}\cdot\frac{TG}{TY}\cdot\frac{RY}{PR}\iff 1=\frac{PB}{PR}\cdot\frac{TG}{TY}\iff 2TG=\frac{PR\cdot BS}{PB}$$By sine theorem $$\frac{PC}{AP}=\frac{\sin\angle ABC}{\sin\angle ACB}=\frac{AC}{AB}.$$Menelaos in $\triangle BQC$ with line $P,S,A$ yields $$1=\frac{CP}{BP}\cdot\frac{BS}{QS}\cdot\frac{QA}{AC}\iff 1=\frac{CP}{BP}\cdot\frac{BS}{BS+2SX}\cdot\frac{AB}{AC}\iff 1=\frac{AP}{BP}\cdot\frac{BS}{BS+2SX}\iff$$$$\frac{2SX}{BS}=\frac{AP-PS}{PS}\iff 2SX=\frac{BS\cdot AS}{PS}$$. Since $PR=AS$ and $PS=PB$ we have finished. 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Take a point B' on extension of BY such that BY=YB'.Then ABPB' is parallelogram and triangles AQB' and ABR are congruent.Rest is easy.

Ismayil_Orucov wrote: Take a point B' on extension of BY such that BY=YB'.Then ABPB' is parallelogram and triangles AQB' and ABR are congruent.Rest is easy. Thanks for Elegant Solution