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

  • 2
What are you looking for?