From Gogeometry Using Area of a triangle Tangents to a circle Inscribed circle of a triangle b=ACAN=AM, AF=AE, MF=NE,CP=CO, CF=CG, PF=OG,BE=BGr=DF=DE=DGx=JM=JN=LO=LPMP=2x S=[ABC]=r(AB+BC+AC)/2 => S=r(AB+CB+b)/2 AB=AE+EB=AF+EBCB=CG+GB=CF+GBAB+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=rEBS=x(AM+MF+CP+PF)+r(MF+PF) + rEB(2) S=xb+2xr + rEB (1) and (2) => S=r(b+BE)=xb+2xr+ rEBbr=xb+ 2xr Therefore x=br/(b+2r)