A circumscribed quadrilateral $ABCD$ is given. $E$ and $F$ are the intersection points of opposite sides of the $ABCD$. It turned out that the radii of the inscribed circles of the triangles $AEF$ and $CEF$ are equal. Prove that $AC \bot BD$.
Source: 239 2008 S2
Tags: geometry
A circumscribed quadrilateral $ABCD$ is given. $E$ and $F$ are the intersection points of opposite sides of the $ABCD$. It turned out that the radii of the inscribed circles of the triangles $AEF$ and $CEF$ are equal. Prove that $AC \bot BD$.