a circle is inscribed in a triangle ABC having sides AB=10 cm AC=8cm BC=12cm such that it touches AB at F AC at E and BC at D.

find AF , BD and CE

Tangents drawn from an external point to a circle are equal.

Let 

AD = AF = x 

BD = BE = y

and CE = CF = z 

 

Now, 

AB = AD + DB = x + y = 10 ...(1)

BC = BE + EC = y + z = 12 ...(2)

AC = AF + FC = x + y = 8 ...(3)

 

Adding (1), (2) and (3), we get,

 

⇒2 (x + y + z) = 30

⇒ (x + y + z) = 15 ...(4)

 

On Solving (1) and (4), we get, z =5

Solving (2) and (4), we get x = 3

Solving (3) and (4), we get y = 7

 

Hence, 

AD = AF= x = 3 cm

BD = BE = y = 7 cm

CE =CF = z = 5 cm

 

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