14. The incircle of triangle ABC touches the sides BC, CA andAB at D, E and F respectively. Prove thatAB+BD+CE = AE+CD+BF
Given: Incircle of the ΔABC touches the sides BC, CA and AB at D, E and F respectively.
To prove: AF + BD + CE = AE + BF + CD
Proof:
We know that, the length of tangent drawn from and external point to a circle are equal.
∴ AF = AE ...(1)
BD = BF ...(2)
CE = CD ...(3)
Adding (1), (2) and (3), we get
AF + BD + CE = AE + BF + CD ...(4)
Hence proved