Point $M$ on side $AB$ of quadrilateral $ABCD$ is such that quadrilaterals $AMCD$ and $BMDC$ are circumscribed around circles centered at $O_1$ and $O_2$ respectively. Line $O_1O_2$ cuts an isosceles triangle with vertex M from angle $CMD$. Prove that $ABCD$ is a cyclic quadrilateral.
(M. Kungozhin)
We omit the trivial case $AB\parallel CD$. Then WLOG the rays $AB$ and $DC$ meet at $S$. By the given condition the segments $DM$ and $CM$ are antiparallels in the angle $S$. Thus $SM^2=SC\cdot SD$. Also because of the same reason
$$\angle SO_1M=\angle O_1MA-\angle O_1SA=90^\circ-\angle O_2DS-\angle O_2SD=\angle SMO_2.$$Hence $SM^2=SO_1\cdot SO_2$. Combining this with our earlier result $SM^2=SC\cdot SD$ we get that $DO_1O_2C$ is cyclic. Thus
$$180^\circ=\angle DO_1O_2+\angle DCO_2=90^\circ+\frac{\angle DAB}{2}+\frac{\angle DCB}{2}$$$\rightarrow$
$$\angle DAB+\angle DCB=180^\circ$$as desired.