goldeneagle wrote: Consider a circle $C_1$ and a point $O$ on it. Circle $C_2$ with center $O$, intersects $C_1$ in two points $P$ and $Q$. $C_3$ is a circle which is externally tangent to $C_2$ at $R$ and internally tangent to $C_1$ at $S$ and suppose that $RS$ passes through $Q$. Suppose $X$ and $Y$ are second intersection points of $PR$ and $[size=200][color=\#BF0040]QR[/color][/size]$ with $C_1$. Prove that $QX$ is parallel with $SY$. Think wrong..... This Statement is correct : Suppose $X$ and $Y$ are second intersection points of $PR$ and $OR$ with $C_1$.

Another solution can be obtain from this observation : $R$ is incenter of $\triangle YPQ$ because $OR=OP=OQ$ .

mahanmath wrote: Another solution can be obtain from this observation : $R$ is incenter of $\triangle YPQ$ because $OR=OP=OQ$ . I used this and this point that $RS$ passes through insimilicenter of $C_1$ and $C_2$

As you see in picture ,line $XS$ and $d$ (tangent at $Q$ to $C_1$) are image of $C_2,C_3$ under inversion with pole $R$ and with radius power of $R$ WRT $C_1$ .Note $C_2 , C_3 $ are tangent , so $d \parallel XS$ and it means $Q$ is midpoint of arc $XQS$ . $\measuredangle QYX = \measuredangle QYS = \measuredangle QXS = \measuredangle QSX \\ = \measuredangle QPX = \measuredangle XPY$ ($R$ is incenter of $\triangle YPQ$ as I mentioned in above post) $=\measuredangle XPY = \measuredangle XQY$ So $\boxed {\measuredangle XQY = \measuredangle QYS}$ $\implies$ $QX \parallel SY$

goldeneagle wrote: Consider a circle $C_1$ and a point $O$ on it. Circle $C_2$ with center $O$, intersects $C_1$ in two points $P$ and $Q$. $C_3$ is a circle which is externally tangent to $C_2$ at $R$ and internally tangent to $C_1$ at $S$ and suppose that $RS$ passes through $Q$. Suppose $X$ and $Y$ are second intersection points of $PR$ and $OR$ with $C_1$. Prove that $QX$ is parallel with $SY$. yes , angle chasing kills the problem!!