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.​

Answer 1

Answer:

Step-by-step explanation:

Given, s is the semi perimeter of ΔABC. AB = c, BC = a and CA = b

A circle touches the sides AB, BC and CA of ΔABC in F, D and E respectively. Therefore, AB, BC and CA are the tangents to the circle.

Let BD = x

∴ CD = a – x

We know that, the lengths of two tangents drawn from an external point to a circle are equal.

∴ BD = DF = x

CD = CE = a – x

and AE = AF

AE = AC – CE = b – (a – x) = b – a + x

AF = AB – BF = c – x

∴ b – a + x = c – x

⇒ 2x = c + a – b ...(1)

Semi perimeter of ΔABC,

$s = \frac{a +b +c }{2}$

$(a + b + c) = 2s$

$ca = 2s - b \: \: \: \: \: \: \: ........(2)$

From (1) and (2), we get

2x = 2s – b – b

∴ 2x = 2s –2b

⇒ x = s – b

Thus, BD = s – b

Answer 2

$\huge\underline{\overline{\mid{\bold{\pink{ANSWER-}}\mid}}}$

Given, s is the semi perimeter of ΔABC. AB = c, BC = a and CA = b

A circle touches the sides AB, BC and CA of ΔABC in F, D and E respectively. Therefore, AB, BC and CA are the tangents to the circle.

Let BD = x

∴ CD = a – x

We know that, the lengths of two tangents drawn from an external point to a circle are equal.

∴ BD = DF = x

CD = CE = a – x

and AE = AF

AE = AC – CE = b – (a – x) = b – a + x

AF = AB – BF = c – x

∴ b – a + x = c – x

⇒ 2x = c + a – b ...(1)

Semi perimeter of ΔABC,

From (1) and (2), we get

2x = 2s – b – b

∴ 2x = 2s –2b

⇒ x = s – b

Thus, BD = s – b