In a $\triangle A B C, D, E$ and $F$ respectively are the points of contact of the incircle with the sides…
In a $\triangle A B C, D, E$ and $F$ respectively are the points of contact of the incircle with the sides $A B, B C$ and $C A$ such that $A D=\alpha, B E=\beta$ and $C F=\gamma$, then $\frac{\alpha \beta \gamma}{\alpha+\beta+\gamma}=$
$R^2$
$2 R$
$2 r$
$r^2$
Solution
From figure
$
\begin{array}{ll}
\operatorname{ar}(\triangle A B C)=\operatorname{ar}(\triangle A O B)+\operatorname{ar}(\triangle B O C)+\operatorname{ar}(\triangle C O A) \\
\Rightarrow \quad & \left.S=\frac{1}{2} c r+\frac{1}{2} a r+\frac{1}{2} b r \text { [where } \operatorname{ar}(\triangle A B C)=S\right] \\
\Rightarrow & S=\frac{1}{2} r(a+b+c) \\
\Rightarrow & S=\frac{1}{2} r(2 s) \quad \text { [where } s \text { is semi perimeter] } \\
\Rightarrow & S=r s \Rightarrow r=\frac{S}{s} \Rightarrow r^2=\frac{s^2}{s^2} \\
\Rightarrow & r^2=\frac{s(s-a)(s-b)(s-c)}{s^2} \\
\Rightarrow & r^2=\frac{\alpha \cdot \beta \cdot \gamma}{\alpha+\beta+\gamma} \quad[\because 2 s=2 \alpha+2 \beta+2 \gamma]
\end{array}
$