In the given figure, ABCD is a square sheet of paper. It is folded along EF such that A goes to a point A′ different from B and C, 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.[asy]
size(5cm);
pair A=(0,0),B=(1,0),C=(1,1),D=(0,1),Ap=(1,0.333),Dp,Ee,F,G;
Ee=extension(A,B,(A+Ap)/2,bisectorpoint(A,Ap));
F=extension(C,D,(A+Ap)/2,bisectorpoint(A,Ap));
Dp=reflect(Ee,F)*D;
G=extension(C,D,Ap,Dp);
D(MP("A",A,W)--MP("E",Ee,S)--MP("B",B,E)--MP("A^{\prime}",Ap,E)--MP("C",C,E)--MP("G",G,NE)--MP("D^{\prime}",Dp,N)--MP("F",F,NNW)--MP("D",D,W)--cycle,black);
draw(Ee--Ap--G--F);
dot(A);dot(B);dot(C);dot(D);dot(Ap);dot(Dp);dot(Ee);dot(F);dot(G);
draw(Ee--F,dashed);
[/asy] geometryPlane GeometryINMOindiainradiussquare