Assume that $ΔABC\sim ΔAB_1C_1$ (where the corners are ordered in that way, i.e. direct similarity). Show that the intersection $M≠A$ between the circles around the two triangles coincides with the intersection between $BB_1$ and $CC_1$
There is a lot of properties for triangles inscribed in circles, some of them are listed here.What i struggle to see is how to use this when considering two different circles.
Let $M$ be the intersection point of $BB_1$ and $CC_1$.
Since the triangles $ABC$ and $AB_1C_1$ are similar we have: $$ \angle BAB_1=\angle BAC_1-\angle C_1AB_1 =\angle CAC_1+\underbrace{\angle CAB-\angle C_1AB_1}_{=0}=\angle CAC_1. $$
Since further: $$ \frac{AB}{AB_1}=\frac{AC}{AC_1} $$ we can conclude: $$ \begin{align} \triangle CAC_1\sim \triangle BAB_1\implies\quad&\\ \\ 1.\quad \angle ABB_1=\angle ACC_1&\implies \angle ABM=\angle ACM\\ &\implies AMBC\text{ cyclic }\implies M\in \bigcirc ABC;\\ \\ 2.\quad \angle AB_1B=\angle AC_1C&\implies \angle AB_1M=\angle AC_1M\\ &\implies AMB_1C_1\text{ cyclic }\implies M\in \bigcirc AB_1C_1.\\ \end{align} $$
In the above proof it was assumed that the point pairs $B,C$ and $B_1,C_1$ are in the different half-planes created by the line $AM$. The same argument can be however applied with minor changes for all possible arrangements: $$(BC|B_1C_1),(BCB_1|C_1),(BCB_1C_1|),(BB_1|CC_1).$$
The second arrangement as an example is shown in the next figure: