A circle touches the sides , , of ΔABC at points D, E, F respectively. BD = x, CE = y, AF = z. Prove that the area of ΔABC = .

Given that a circle touches the sides BC, CA and AB of ΔABC at points D, E and F respectively.

Let BD = x, CE = y and AF = z.

We have to prove that area of ΔABC =

Proof:

BD and BF are tangents drawn from B. And D and F are points of contact.

∴ BD = BF = x

Similarly for tangents CE and CD drawn by C and tangents AE and AF drawn from A,

CE = CD = y and AF = AE = z

Sides of ΔABC,

⇒ AB = c = AF + BF = z + x … (1)

⇒ BC = a = BD + DC = x + y … (2)

⇒ CA = b = CE + AE = y + z … (3)

In ΔABC, 2s = AB + BC + AC

= (z + x) + (x + y) + (y + z)

= 2 (x + y + z)

∴ s = x + y + z … (4)

We know that area of ΔABC =

Area =

=

=

Hence proved.