# A circle touches Given that a circle touches all the three sides of a right angled ΔABC.

We have to prove that radius of a circle = Proof:

Let radius be r and the centre of the circle touching all the sides of a triangle be I i.e. incentre.

In the figure,

ID = IE = IF = r

Since ΔABC is a right angled triangle, B = 90°.

Also ID BC and AB BC.

ID || AB and ID || FB

Similarly, IF || BD

IFBD is a parallelogram.

FB = ID = r and BD = IF = r … (1)

Parallelogram IFBD is a rhombus.

Since B = 90°, parallelogram IFBD is a square.

Now AE = AF

AE = AB – FB

= AB – r [From (1)] … (2)

And CE = CD

CE = BC – BD

= BC – r [From (1)] … (3)

Now, AC = AE + CE,

AC = AB – r + BC – r

AC = AB + BC – 2r

2r = AB + BC – AC

r = The radius of a circle is .

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