Question

In the given figure, the circle touches the sides of $△ABC$ $BC$, $CA$ and $AB$ at $D$, $E$ and $F$ respectively.

Show that $AF+BD+CE=AE+BF+CD=\frac{1}{2}\left(Perimeterof△ABC\right)$

Answer 1

Since lengths of the tangents from an exterior point to a circle are equal.

$AF=AE$
$BD=BF$
$CE=CD$

Adding equations $\left(i\right),\left(ii\right)$ and $\left(iii\right)$, we get
$AF+BD+CE=AE+BF+CD$
Now,
Perimeter of $△ABC=AB+BC+AC$
$\Rightarrow$ Perimeter of $△ABC=(AF+FB)+(BD+CD)+(AE+EC)$
$\Rightarrow$ Perimeter of $△ABC=(AF+AE)+(BF+BD)+(CD+CE)$
$\Rightarrow$ Perimeter of $△ABC=2AF+2BD+2CE$
$\Rightarrow$ Perimeter of $△ABC=2(AF+BD+CE)$
$\Rightarrow AF+BD+CE=\frac{1}{2}\left(Perimeterof△ABC\right)$
Hence, $AF+BD+CE=AE+BF+CD=\frac{1}{2}\left(Perimeterof△ABC\right)$