Question

Let's denote the semi-perimeter of a triangle ABC in which BC$=$a, CA$=$b, AB$=$c. If a circle touches the sides BC, CA, AB at D, E, F respectively, prove that BD$=$s$-$b.

Answer 1

Let $BD=x$ as shown in the figure.

$BD=BF=x$ [ Tangents drawn from the same point $B$ to the circle. ]

Similarly, $CD=CE=z$ and $AF=AE=y$

Perimeter of $△ABC=AB+BC+AC$

We know, $s$ is semi- perimeter then,

$\Rightarrow$ $2s=2(x+y+z)$

$\Rightarrow$ $s=x+y+z$

$\Rightarrow$ $s-(y+z)=x$

$\Rightarrow$ $s-AC=x$

$\Rightarrow$ $s-b=x$ [ Since, $AC=b$ ]

$\therefore$ $BD=s-b$ ---- Hence proved.