Login/Sign Up
Home » 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.
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.
CBSE | Circles | Class 10 | Exercise 9.4 | Maths | NCERT Exemplar
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.
NCERT Exemplar Class 10 Maths Chapter 9 Ex. 9.4 Question 2

As indicated by the inquiry,

A triangle ABC with BC = a , CA = b and AB = c . Likewise, a circle is engraved which contacts the sides BC, CA and AB at D, E and F individually and s is semi-border of the triangle

To Prove: BD = s – b

Evidence:

As per the inquiry,

We have,

Semi Perimeter = s

Border = 2s

    \[2s\text{ }=\text{ }AB\text{ }+\text{ }BC\text{ }+\text{ }AC\text{ }\left[ 1 \right]\]

As we probably are aware,

Digressions drawn from an outside highlight a circle are equivalent

So we have

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

BF = BD [3] [Tangents From point B]

Compact disc = CE [4] [Tangents From point C]

Adding [2] [3] and [4]

    \[AF\text{ }+\text{ }BF\text{ }+\text{ }CD\text{ }=\text{ }AE\text{ }+\text{ }BD\text{ }+\text{ }CE\]

Stomach muscle + CD = AC + BD

Adding BD both side

Abdominal muscle + CD + BD = AC + BD + BD

Stomach muscle + BC – AC = 2BD

Stomach muscle + BC + AC – AC – AC = 2BD

2s – 2AC = 2BD [From 1]

2BD = 2s – 2b [as AC = b]

BD = s – b

Consequently Proved.

i

More Material