Blastoor wrote: Circle $\omega$ with centre $M$ and diameter $XY$ is given. Point $A$ is chosen on $\omega$ so that $AX<AY$. Points $B, C$ are chosen on segments $XM, YM$, respectively, in a way that $BM=CM$. A parallel line to $AB$ is constructed through $C$; the line intersects $\omega$ at $P$ so that $P$ lies on the smaller arc $\widehat{AY}$. Similarly, a parallel line to $AC$ is constructed through $B$; the line intersects $\omega$ at $Q$ so that $Q$ lies on the smaller arc $\widehat{XA}$. Lines $PQ$ and $XY$ intersect at $S$. Prove that $AS$ is tangent to $\omega$. $T$ is the reflection of $A$ through $XY$. $QB$ intersects $PC$ at $G$. Then we can see $ABGC$ is a parallelogram. Thus, $G$ is the antipode of $A$ wrt $\omega$. Now, we can see that $G(QP,AT)=G(BC,MT)=-1$, then $APTQ$ is harmonic. This yields $PQ$ goes through intersection of tangents of $\omega$ at $A,T$. But $AXTY$ is harmonic, too, we can see that $AS$ is tangent to $\omega$. $\square$

Attachments: