Question

In $\Delta ABC,$ if the incircle touch the sides $BC,CA,AB$ at $D,E,F$, respectively, then $BD+CE+AF$ equals

• A. $s$
• B. $2s$
• C. $\frac{s}{2}$
• D. None of these

Answer 1

The correct option is A $s$

We know that the length of tangent drawn from an external point to a circle are equal.

$\therefore AF=AE$------------(i)

$BD=BF$------------(ii)

$CE=CD$------------(iii)

Add (i), (ii), and (iii), we get

$AF+BD+CE=AE+BF+CD$------------(iv)

Perimeter of $△ABC=AB+BC+CA$

$=(AF+BF)+(BD+CD)+(CE+AE)$

$=(AF+BD)+(BD+CE)+(CE+AF)$-------from (i), (ii), and (iii)

$=2(AF+BD+CE)$

$AF+BD+CE=\frac{1}{2}\left(Perimeterof△ABC\right)$

We know that $s$ is the semi-perimeter.

$AF+BD+CE=s$