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.
Answers
Answer:
According to the question,
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 semiperimeter of the triangle
To Prove: BD = s – b
Proof:
According to the question,
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
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