AustriaMO2013

We denote with $M_{XY}$ the mid-point of the line segment $XY$. Using the intercept theorem, we deduce that $M_{AD}{M}_{BD}$ is parallel to $AB$. $M_{AC}{M}_{BC}$ is parallel to $AB$. $M_{AC}{M}_{AD}$ is parallel to $CD$. $M_{BC}{M}_{BD}$ is parallel to $CD$. Therefore, $M_{AD}{M}_{BD}$ is parallel to $M_{AC}{M}_{BC}$ and $M_{AC}{M}_{AD}$ is parallel to $M_{BC}{M}_{BD}$. Furthermore, $M_{AC}{M}_{BC}$ is orthogonal to $M_{AC}{M}_{AD}$, since $CD$ is orthogonal to $AB$. Therefore, the four mid-points form a rectangle. $\square$