Let $P \equiv AB \cap A'B'$ and $M \equiv AA' \cap BB'.$ By Menelaus' theorem for $\triangle PBB'$ cut by $\overline{AA'M},$ keeping in mind that $\overline{AB} \cong \overline{A'B'}$ and $\overline{PA} \cong \overline{PA'},$ we get $\frac{BM}{MB'}=\frac{A'P}{B'A'} \cdot \frac{AB}{AP}=\frac{AP}{AP} \cdot \frac{AB}{AB}=1$ $\Longrightarrow$ $M$ is the midpoint of $\overline{BB'}.$

Let O be the centre of the circle and AA' meet BB' at M. We have <AOA'=2<BAA' but since the line segment AB is rotated to A'B' about centre O, <AOA'=<BOB' and hence, <BOB'=2<BAA'. Consider triangle OBB'. It is isosceles as OB=OB'. We already know <BOB'=2<BAA' and so, the remaining angles <OBB', <OB'B are both 90-<BAA'. If we consider triangle AOB, <OAB is 90 degrees as AB is tangent to the circle (O) and so, <ABO=90-<AOB. Take triangle ABM. <MAB=<BAA'(they are essentially the same angle), <ABM=<ABO+<OBM=90-<AOB+<90-<BAA'(as derived earlier). So, we can find <AMB. We see that it is equal to <AOB which proves that the quadrilateral AOMB is cyclic. Now, one of the angles of the quadrilateral <OAB is 90 degrees and hence, so is <OMB or in other words, OM is perpendicular to AB. Now, in isosceles triangle OBB', M lies on BB' and OM is perpendicular to BB' implies that M is the midpoint of BB'.

Take the reflections of $B$ in $A$ and respectively $B'$ in $A'$, the problem is obvious then. Best regards, sunken rock

centred at O: $A \sim A',B \sim B'$. Take $M=AA' \cap BB'$ , $\triangle OAA' \sim \triangle OBB',$ $\quad MOBA$ is cyclic.

Let $D = \overline{AA'} \cap \overline{BB'}$. Because $C$ is the center of the spiral similarity $\overline{AA'} \to \overline{BB'}$, it is also the Miquel Point of $AA'B'B$, so $CA'B'D$ and $CADB$ are cyclic. Then it suffices to show that $$\angle BCD = \angle B'CD \iff \angle BAD = \angle AA'B'$$as $CB = CB'$, but this expression is obviously true by equal tangents.

This was in Titu’s book 106 geometry problems.