Question

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

Answer 1

We know that tangent drawn from external point to a circle are equal

So $AD=AF=x$

Or $BD=BE=y$

And $CF=CE=z$

Now, $AB=AD+DB=x+y=12$ cm ...(1)

Or $BC=BE+EC=y+z=8$ cm ...(2)

And $AC=AF+CF=x+z=10$ cm ...(3)

Adding the above three equation, we get

$2(x+y+z)=12+8+10=30$ cm

Or $x+y+z=15$ cm

As per equation (1),

$x+y=12$ cm

Then, $z=15-10=5$ cm $=CF$

As per equation (3),

$x+z=12$ cm

Then, $y=15-12=3$ cm $=BE$

As per equation (2),

$y+z=8$ cm

Then, $x=15-8=7$ cm $=AD$