MathDB

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

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]

Problem Statement