Math, asked by popo0000, 9 months ago

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

Answered by Skyllen
7

[HeY Mate]

Answer:

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.

Solution:

Given,

  • S is semi perimeter of triangle.
  • In ABC, 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.

To Prove: BD = s – b

Proof:

Semi Perimeter = s

Perimeter = 2s

2s = AB + BC + AC [1]

Tangents drawn from an external point to a circle are equal...

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!

I Hope It Helps You✌️

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Answered by Anonymous
4

Answer:

Proof:

Semi Perimeter = s

Perimeter = 2s

2s = AB + BC + AC [1]

【Tangents drawn from an external point to a circle are equal】...

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

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