The incircle of `DeltaABC` touches side BC at D. The difference between BD and CD (R is circumradius of `DeltaABC`) is

The incircle of Delta ABC touches the sides AB, BC and CA of the triangle at the points D, E and F. Prove that AD + BE +CF=AF+CE +BD=1/2 (The perimeter of DeltaABC )

In DeltaABC , the incircle touches the sides BC, CA and AB, respectively, at D, E,and F. If the radius of the incircle is 4 units and BD, CE, and AF are consecutive integers, then

In DeltaABC , the incircle touches the sides BC, CA and AB, respectively, at D, E,and F. If the radius of the incircle is 4 units and BD, CE, and AF are consecutive integers, then

In the given figure, D, E and F are the points where the incircle of the DeltaABC touches F the sides BC, CA and AB respectively. Show that AF+ BD + CE = AE + BF + CD = (perimeter of Delta ABC)