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If $\alpha, \beta, \gamma$ are the lengths of the tangents from the vertices of a triangle to its incircle. Then
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Verified Answer
The correct answer is:
$\alpha+\beta+\gamma=\frac{1}{r^2}(\alpha \beta \gamma)$
It is given that $\alpha, \beta, \gamma$ are the length of tangents from the vertices of a triangle to its incircle.
Semi-perimeter of $\triangle A B C$,
$$
\begin{aligned}
\Rightarrow \quad S=\alpha+\beta & +\gamma \\
\text { Area of } \triangle A B C & =\sqrt{s(s-a)(s-b)(s-c)} \\
& =\sqrt{(\alpha+\beta+\gamma)(\alpha)(\beta)(\gamma)}
\end{aligned}
$$
As we know,
$$
\begin{aligned}
& r=\Delta / s \\
& \because \quad r=\frac{\sqrt{(\alpha+\beta+\gamma)(\alpha \beta \gamma)}}{\alpha+\beta+\gamma} \\
& \Rightarrow \quad r^2=\frac{(\alpha+\beta+\gamma)(\alpha \beta \gamma)}{(\alpha+\beta+\gamma)^2} \\
& \Rightarrow \text { Hence, } \alpha+\beta+\gamma=\frac{\alpha \beta \gamma}{r^2}
\end{aligned}
$$
Semi-perimeter of $\triangle A B C$,
$$
\begin{aligned}
\Rightarrow \quad S=\alpha+\beta & +\gamma \\
\text { Area of } \triangle A B C & =\sqrt{s(s-a)(s-b)(s-c)} \\
& =\sqrt{(\alpha+\beta+\gamma)(\alpha)(\beta)(\gamma)}
\end{aligned}
$$
As we know,
$$
\begin{aligned}
& r=\Delta / s \\
& \because \quad r=\frac{\sqrt{(\alpha+\beta+\gamma)(\alpha \beta \gamma)}}{\alpha+\beta+\gamma} \\
& \Rightarrow \quad r^2=\frac{(\alpha+\beta+\gamma)(\alpha \beta \gamma)}{(\alpha+\beta+\gamma)^2} \\
& \Rightarrow \text { Hence, } \alpha+\beta+\gamma=\frac{\alpha \beta \gamma}{r^2}
\end{aligned}
$$
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