Two circles A and B are touching and have a common tangent which meets A at M and B at N. Let MP be a diameter of A and let the tangent from P to B touch it at Q Show that MP=PQ
I've drawn it out and a few different lines through the shape to try to get a new perspective, labelled all the angles and sides but didn't get anywhere... would really appreciate the help, I've been trying a few hours nowdiagram
Proof sketch with inversion: Invert through $P$ with radius $PM$, and denote the inversions with $'$. Clearly $P=P'$ and $M=M'$, and write the common tangent as $l$. Now, $l' = A$ (and $A'=l$), so consider $B'$. In particular, $B'$ must lie tangent to $A'=l$ and $l'=A$, and to $PQ'$. We don't yet know where $Q'$ lies, but it is certainly on the line $PQ$, so $B'$ is tangent to $l$, $A$ and $PQ$. But $B$ is tangent to these, and it is not hard to see now that $B=B'$. Furthermore, $Q=Q'$, so $PM = PQ$, as required.
The important idea here is that the whole diagram is invariant under this particular inversion.
