Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

Invert with respect to a circle centred at $A$ and denote by $X^*$ the image of a point $X \neq A$ under this inversion. The circles $\gamma$ and $\gamma'$ invert into straight lines $B^*C^*$ and $B^*C'^*$, and the tangents at $A$ into lines through $A$, parallel to $B^*C^*$ and $B^*C'^*$. The line $CC'$ inverts into the circle $AC^*C'^*$, and 'CC' separating $A$ and $B'$ is equivalent to '$B'$ lying inside circle $AC^*C'^*$. So $AC^*B^*C'^*$ is a parallelogram, obtuse-angled at $A$ and $B^*$. Draw the line through $A$, parallel to $C^*C'^*$, meeting the lines $B^*C^*$ and $B^*C'^*$ at $D$ and $D'$, respectively. Then $A$, $C^*$ and $C'^*$ are the midpoints of the sides of the triangle $B^*DD'$, and the line $DD'$ is the inverse of the line through $A$ on which $\gamma$ and $\gamma'$ intercept equal segments. The circle $AC^*C'^*$ is the nine-point circle of the triangle $B^*DD'$; by Feuerbach's theorem, it touches the incircle of that triangle. Since this incircle is the inverse of the circle $\Gamma$ in the original configuration, touching $\gamma$ and $\gamma'$ externally and the line $CC'$, the conclusion follows.