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 =
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 =
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