Any solutions anyone?

parmenides51 wrote: Let $P,Q,R$ be three points on a circle $k_1$ with $|PQ|=|PR|$ and $|PQ|>|QR|$. Let $k_2$ be the circle with center in $P$ that goes through $Q$ and $R$. The circle with center $Q$ through $R$ intersects $k_1$ in another point $X\ne R$ and intersects $k_2$ in another point $Y\ne R$. The two points $X$ and $R$ lie on different sides of the line through $PQ$. Show that the three points $P$, $X$, $Y$ lie on a common line. Let's call the circle with center $Q$ and radius $QR$, $k3$. Then because of $QR$ and $QX$ are radius of $k3$, $QR=QX$. And because of $QRPX$ cyclic, we get PQ is bisector of $\angle RPX$. Let $\angle RPX = \alpha \Longrightarrow \angle RPQ = \alpha /2$. Because of $QR = QY \Longrightarrow$ small arcs $\overarc{QR} = \overarc{QY}$(Edit: in $k2$) $ \Longrightarrow \angle RPQ = \angle QPY$. Then, $\angle RPY = \angle RPQ + \angle QPY = 2 \angle RPQ = \alpha = RPX$.So $\angle RPX = \angle RPY \Longrightarrow P,X,Y$ collinear.

Let $\angle PQR=a$. Then $\angle PXR=a$ and $\angle RXY=180^{\circ}-\frac{1}{2}\angle RQY=180^{\circ}-a$, as desired.