Point $X$ is on the circumference of the circle $PQR$ and $PY$ is a perpendicular on $XR$. Finding the value of $QX + XR$ 
$\triangle PQR$ is an isosceles triangle where $PQ = PR$. Point $X$ is on the cicumcircle of $\triangle PQR$ such that it is in the opposite region of $P$ with respect to $QR$. $PY$ $\perp$ $XR$ and $XY$ = $12$. What is the value of $QX  + XR$?


I am unable to solve the problem because I couldn't use the condition of $XY$ = $12$ and $\angle PYX$ = $90^\circ$ $\triangle PQR$ being an isosceles triangle. Then how can I suppose to get the value of $QX + XR$?
A small hint will be enough for me to proceed.
Source: Bangladesh Math Olympiad
 A: 
Choose a point $R'\in [XY]$ such that $\color{red}{YR=YR'}$ $\Rightarrow PR'=PR=PQ\Rightarrow \angle PR'Q=\angle R'QP$.
Since the quadrilateral $PRXQ$ is cyclic: $$\angle PRY=\angle YR'P=180°-\angle XQP \Rightarrow \angle PR'X=\angle XQP$$
$$\because \angle PR'Q=\angle R'QP \Rightarrow \angle XQR'=\angle QR'X$$
$$\therefore \color{red}{XR'=XQ}\Rightarrow QX+XR=XR'+XY+R'Y=2XY=24$$
A: Let $\measuredangle PRQ=\alpha$ and $\measuredangle XRQ=\beta$.
Thus, $$PX=\frac{12}{\cos\alpha}$$ and by the law of sines for $\Delta QPX$ and for $\Delta PRX$ we obtain:
$$\frac{QX}{\sin\beta}=\frac{\frac{12}{\cos\alpha}}{\sin(\alpha+\beta)}$$ and
$$\frac{RX}{\sin(2\alpha+\beta)}=\frac{\frac{12}{\cos\alpha}}{\sin(\alpha+\beta)}.$$
Id est, $$QX+RX=\frac{12\sin(2\alpha+\beta)+12\sin\beta}{\cos\alpha\sin(\alpha+\beta)}=\frac{24\sin(\alpha+\beta)\cos\alpha}{\cos\alpha\sin(\alpha+\beta)}=24.$$
A: Drop A perpendicular from $P$ onto $QR$ at $S$, so $QR=2\cdot SR$. And let $PR=PQ=a$
Note that $\angle PXR=\angle PXQ=\angle PRS=\alpha$
Now from Ptolemy theorem in $PQXR$:
$$QX\cdot PR+XR\cdot PQ=PX\cdot QR$$ $$a\cdot(QX+XR)=(12\sec\alpha)\cdot 2a\cos\alpha\\QX+XR=24$$
