Question

Let s denote the semi-perimeter of a triangle ABC in which BC = a, CA = b, AB = c. If a circle touches the sides BC, CA, AB at D, E, F, respectively, prove that BD = s – b.

Answer 1

A triangle ABC with BC = a, CA = b and AB = c.

Also, a circle is inscribed, which touches the sides BC, CA and AB at D, E and F, respectively and s is semi-perimeter of the triangle

To Prove: BD = s – b

Proof: 

We have,

Semi Perimeter = s

Perimeter = 2s

2s = AB + BC + AC  ...[1]

As we know,

Tangents drawn from an external point to a circle are equal

So we have

AF = AE  ...[2] [Tangents from point A]

BF = BD  ...[3] [Tangents from point B]

CD = CE  ...[4] [Tangents from point C]

Adding [2] [3] and [4]

AF + BF + CD = AE + BD + CE

AB + CD = AC + BD

Adding BD both side

AB + CD + BD = AC + BD + BD

AB + BC – AC = 2BD

AB + BC + AC – AC – AC = 2BD

2s – 2AC = 2BD  ...[From 1]

2BD = 2s – 2b  ...[As AC = b]

BD = s – b

Hence proved.