Let s denotes the semi-perimeter of a ΔABC in which BC = a, CA = b and AB = c. If a circle touches the sides BC, CA, AB at D, E, F, respectively. Prove that BD = s – b.

Given : 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 :

Given that

Semi Perimeter = s

Perimeter = 2s

Implies that

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 !

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