Two noncongruent circles intersect at $X$ and $Y$. Their common (external) tangents intersect at $Z$. One of the common tangents touches the circles at $P$ and $Q$. Prove that $ZX$ is tangent to the circumcircle of triangle $PXQ$.
I let $M$ be a point that is the intersection of XZ and the smaller circle, connecting $Q$ and $M$ gives us trapezium $XMQP$ where $XP||MQ$. But now I'm stuck what should I do next? I know I can move forward with angle chasing but I don't see how.
Best Answer
By the power of the point $Z$: $$ZQ^2 = ZX\cdot ZM$$
As you mentioned we have $${ZM\over ZX} = {ZQ\over ZP}\implies ZM = ZX\cdot {ZQ\over ZP}$$
so $$ZQ^2 = ZX^2\cdot {ZQ\over ZP} \implies ZX^2 = ZP\cdot ZQ$$
and we are done.