Question

8
8)
Given : A circle inscribed in a right
angled AABC. If LACB = 90° and the
radius of the circle is r.
To prove: 2 r= a + b-c​

Answer 1

Concept Used :-

• 1. Radius and tangent are perpendicular to each other.

• 2. Length of tangent drawn to a circle from external point are equal.

$\large\underline{\sf{Solution-}}$

Given that,

A circle with centre O and radius r is inscribed in a right angle triangle ABC right-angled at C.

• AB = c units

• BC = a units

• CA = b units

Let circle touches the three sides AB, BC and CA at D, E and F respectively.

Now,

• OE is radius of incircle

and

• BEC is tangent to incircle at point E.

We know that, radius and tangent are perpendicular to each other.

$\bf\implies \:OE \: \perp \: BC - - - (1)$

Similarly,

• OF is radius of incircle

and

• AFC is tangent to incircle at point F.

$\bf\implies \:OF \: \perp \: AC - - - (2)$

Also,

$\rm :\longmapsto\: \angle \: ACB \: = \: 90 \degree$

Now,

In quadrilateral OECF,

$\rm :\longmapsto\: \angle \: OEC \: = \: 90 \degree$

$\rm :\longmapsto\: \angle \: CFO \: = \: 90 \degree$

$\rm :\longmapsto\: \angle \: FCE \: = \: 90 \degree$

So,

$\rm :\longmapsto\: \angle \: EOF \: = \: 90 \degree$

$\bf\implies \:OECF \: is \: a \: rectangle$

$\rm :\implies\:OE = CF = r$

and

$\rm :\implies\:OF = EC = r$

Now,

$\rm :\longmapsto\:BE = BC - EC = a - r$

Now,

We know,

Length of tangents drawn from external point to a circle are equal.

$\rm :\implies\:BD = BE = a - r - - (3)$

Also,

$\rm :\longmapsto\:AF = AC - FC = b - r$

So,

$\rm :\longmapsto\:AD = AF = b - r - - - (4)$

Now,

$\rm :\longmapsto\:AB = AD + BD$

$\rm :\longmapsto\:c = a - r + b - r$

$\rm :\longmapsto\:c = a - 2r + b$

$\rm :\longmapsto\:2r = a + b - c$

${\boxed{\boxed{\bf{Hence, Proved}}}}$