Question

A circle touches the sides BC, CA, AB of ΔABC at points D, E, F respectively. BD = x, CE = y, AF = z. Prove that the area of ΔABC = √ xyz (x + y +z).

Answer 1

A circle touches the sides BC , CA and AB of ∆ABC at points D, E , F respectively.
Let BD = x , CE = y , AF = z .
we have to prove that area of ∆ABC =
$\bf{\sqrt{xyz(x+y+z)}}$

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 ….....(i)
⇒ BC = a = BD + DC = x + y …....(ii)
⇒ CA = b = CE + AE = y + z ….....(iii)
In ΔABC, 2s = AB + BC + AC
= (z + x) + (x + y) + (y + z)
= 2 (x + y + z)
∴ s = x + y + z ….......(iv)
We know that area of ΔABC = 
$\bf{\sqrt{s(s-a)(s-b)(s-c)}}$
from eqs. (i), (ii), (iii) and (iv),
ar(∆ABC) = $\bf{\sqrt{(x+y+z)(x+y+z-x-y)(x+y+z-y-z)(x+y+z-z-x)}}$
=$\bf{\sqrt{(x+y+z)(x)(y)(z)}}$
=$\bf{\sqrt{xyz(x+y+z)}}$

hence, area of ∆ABC =$\bf{\sqrt{xyz(x+y+z)}}$
hence proved.

Answer 2

Step-by-step explanation:

here is your answer.

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