Question

In $\Delta ABC$, if the incircle touch the sides. $BC,CA,AB$ at $D,E,F$ respectively, then $BD+CE+AF$ equals to: ($s=$ semi perimeter of $\Delta ABC$)

• A. $s$
• B. $3s$
• C. $2s$
• D. $\frac{s}{2}$

Answer 1

The correct option is A $s$

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

$\therefore AF=AE$ ...(1)

$BD=BF$ ...(2)

$CE=CD$...(3)

Adding (1), (2) and (3), we get

$AF+BD+CE=AE+BF+CD$...(4)

Perimeter of $△ABC$

$=AB+BC+CA$

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

$=(AF+BD)+(BD+CE)+(CE+AF)$ [Using (1), (2) and (3)]

$=2(AF+BD+CE)$

$\therefore AF+BD+CE=\frac{1}{2}$(Perimeter of $\Delta ABC$) ...(5)

From (4) and (5), we get

$AF+BD+CE=AE+BF+CD=\frac{1}{2}$ (Perimeter of $\Delta ABC$)$=s$

Answer (A)