Question

In the given figure D, E and F are the points where the incircle of the ∆ABC touches the sides BC, CA and AB respectively. Show that: AF+BD+CE=AE+BF+CD=1/2(perimeter of ∆ABC)

Answer 1

as tangents from same external point are equal
therefore AF = AE, BF = BD, CD = CE
therefore AF+BD+CE=AE+BF+CD
perimeter of ∆ABC = AB + BC + CA
= AF + AE + BD + BF + CD + CE
therefore ½ perimeter = AF+BD+CE = AE+BF+CD