In the given figure, $$ABCD$$ is a square paper. It is folded along EF such that A goes to a point $$A' \in C,B$$ 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 $$A'BE$$.