Question 96
In the given figure, $ABCD$ is a square paper. It is folded along $EF$ such that $A$ goes to a point $A'$ ($A' \neq C, B$) on the side $BC$ and $D$ goes to $D'$. The line $A'D'$ cuts $CD$ in $G$. Show that the inradius of the triangle $GCA'$ is the sum of the inradii of the triangles $GD'F$ and $A'BE$.
