From Gogeometry

Using

• Area of a triangle
• Tangents to a circle
• Inscribed circle of a triangle

b=AC
AN=AM, AF=AE, MF=NE,CP=CO, CF=CG, PF=OG,BE=BG
r=DF=DE=DG
x=JM=JN=LO=LP
MP=2x

S=[ABC]=r(AB+BC+AC)/2 => S=r(AB+CB+b)/2

AB=AE+EB=AF+EB
CB=CG+GB=CF+GB
AB+CB=AF+CF+EB+GB=b+2BE
=> (1) S=r(b+BE)

But S=[AMJN]+[MFDENJ]+[PFDGOLP]+[EBGD]
S=xAM+(r+x)MF+xCP+(r+x)PF + rEB, knowing that [EBGD]=(rEB+rBG)/2=rEB
S=x(AM+MF+CP+PF)+r(MF+PF) + rEB
(2) S=xb+2xr + rEB

(1) and (2) => S=r(b+BE)=xb+2xr+ rEB
br=xb+ 2xr

Therefore x=br/(b+2r)