Question

In triangle ABC angle C = 90degree. A circle of radius arebis inscribed in triangle. a,b,c are the length of side BC, AC and AB. then prove that 2r = a + b - c.​

Answer 1

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

Given that,

A right angle triangle ABC right-angled at C such that BC = a, CA = b and AB = c.

A circle with centre O and radius r is inscribed in a circle touches the sides AB, BC, CA at E, D, F respectively.

Join OD and OE.

We know, Radius and tangent are perpendicular to each other.

So, OD is perpendicular to BC and OE is perpendicular to AC.

So, in quadrilateral ODCE, we have

∠ODC = 90°

∠OEC = 90°

∠ECD = 90°

So, it means ∠DOE = 90°

[ Angle sum property of quadrilateral ]

So, it means ODCE is a rectangle.

So, OD = EC = r and OE = CD = r

Now, BD = BC - CD = a - r

As, we know, length of tangents drawn from external point are equal.

Now, BD and BF are tangents drawn from external point B

$\rm\implies \:\boxed{ \rm{ \:BF=BD=a - r \: }} - - - (1)$

Also, AE = AC - CE = b - r

As, length of tangent drawn from external point are equal.

Now, AE and AF are tangents drawn from external point A.

$\rm\implies \:\boxed{ \rm{ \:AE=AF=b - r \: \: }} - - - (2)$

Now, it is given that,

$\rm \: AB=c$

$\rm \: AF+FB = c$

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

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

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

Hence, Proved