In fig., a circle is inscribed in triangle ABC touches its sides AB, BC and AC at points D, E and F respectively. If AB=12cm, BC=8cm and AC=10cm, then find the length of AD, BE and CF.

We know that tangent drawn from an external point to a circle are equal.
So AD=AF=x
Or BD=BE=y
And CF=CE=z
Now, AB=AD+DB=x+y=12cm .. (1)
Or BC=BE+EC=y+z=8cm .. (2)
And AC=AF+CF=x+z=10cm .. (3)
Adding the above three equation, we get
2(x+y+z)=12+8+10=30cm
Or x+y+z=15cm
As per equation (1), x+y=12cm
Then, z=15-12=3cm=CF
As per equation (3), x+z=10cm
Then, y=15-10=5cm=BE
As per equation (2), y+z=8cm
Then, x=15-8=7cm=AD