Question

The centre of a circle is $O$. $△ABC$ is drawn circumscribing this circle whose sides $BC,CA$ and $AB$ touch the circle at point $D,E$ and $F$ respectively. Prove that
$AF=$ semi-perimeter of $△ABC-BC$.

Answer 1

Let the centre of the circle be $O$.

Let $\Delta ABC$ be drawn circumscribing this circle whose sides $BC,CA,AB$ touch the circle at $D,E,F$ respectively. Therefore $AB,BC,CA$ are tangents to the circle.

Let $AB=a,BC=b,CA=c$.

Let $AF=x$

$\therefore BF=a-x$

We know that the lengths of two tangents drawn from external point to a circle are equal.

$\therefore AF=AE=x$ and $BF=BD=a-x$

Also, $CD=CE$

We know $CD=BC-BD=b-(a-x)=b-a+x$ and $CE=CA-AE=c-x$

$\therefore b-a+x=c-x$

$\Rightarrow 2x=c+a-b...\left(1\right)$

Let $s$ be the semi-perimeter of $\Delta ABC$.

$\therefore s=\frac{a+b+c}{2}$

$\Rightarrow a+b+c=2s$

$\Rightarrow c+a=2s-b...\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we get

$2x=2s-b-b$

$\Rightarrow 2x=2s-2b$

$\Rightarrow x=s-b$

$\Rightarrow AF=semiperimeterof\Delta ABC-BC$