Question

In following figure , the incircle of Δ ABC , touches the sides BC , CA and AB at D , E and F respectively. Show AF + BD + CE = AE + BF + CD

Answer 1

To prove:- AF + BD + CE = AE + BF + CD 

Proof:- AF = AE ----(1) {Length of tangents drawn from an external point to a circle are equal } 

BD = BF ----(2}  

CE = CD ----(3} 

Adding (1), {2} and {3} 

AF + BD + CE = AE + BF + CD