Q. 24.0( 24 Votes )

Let s denotes the

Answer :



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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