29° Olimpiada Matemática del Cono Sur

Let $E$ and $F$ be points in the prolongations of $AB$ and $CD$ such that $DF=DA=CR$ and $BE=BC=AS$, as in the picture. Note that $P$ is the midpoint of $FC$. Then, $MP$ is a midsegment of the triangle $AFC$, which implies that $A\hat{F}C=M\hat{P}C$. Since the triangle $ADF$ is isosceles, we have that $D\hat{A}F=D\hat{F}A=M\hat{P}C$, and then, $A\hat{D}C=2\cdot M\hat{P}C$ (external angle). Similarly, $A\hat{B}C=2\cdot M\hat{Q}A$. Thus, $$A\hat{D}C+A\hat{B}C=2\cdot (M\hat{P}C+M\hat{Q}A)=2\cdot 90^{\circ }=180^{\circ },$$ and, therefore, the quadrilateral $ABCD$ is cyclic.