Show that af+bd+ce=ae+bf+cd= 1/2 , Mathematics

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

Ans: Since the length of tangents from an external point to are equal

∴AF = AE

FB = BD

EC = CD

Perimeter of ΔABC = AB + BC+ AC

= AF + FB + BD + DC + AE + EC

= AF + BD + BD + CE + AF + CE (Θ AF=AE, FB=BD, EC=CD)

= AF + AF + BD + BD + CE + CE

Perimeter of ΔABC = 2(AF + BD+ CE)

∴AF + BD + CE =

1/2 (perimeter of ΔABC) ........(1)

Perimeter of ?ABC = AB + BC + AC

= AF + FB + BD + DC + AE + EC

= AE + BF + BF + CD + AE + CD (Θ AF = AE, FB = BD, EC = CD)

= AE + AE + BF + BF + CD + CD

Perimeter of ΔABC = 2(AE + BF + CD)

∴AE + BF + CD =

1/2 (perimeter of ΔABC) ........(2)

From (1) and (2)

AF + BD + CE = AE + BF + CD = 1/2 (perimeter of ΔABC)

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