If $\alpha, \beta, \gamma$ are the lengths of the tangents from the vertices of a triangle to its incircle.…

If $\alpha, \beta, \gamma$ are the lengths of the tangents from the vertices of a triangle to its incircle. Then
  1. $\alpha+\beta+\gamma=\frac{1}{r^2}(\alpha \beta \gamma)$
  2. $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=r(\alpha \beta \gamma)$
  3. $\alpha+\beta+\gamma=\frac{1}{r}(\alpha \beta \gamma)$
  4. $\alpha^2+\beta^2+\gamma^2=\frac{2}{r}(\alpha \beta \gamma)$

Solution

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} $

Asked in: AP EAMCET 2017 (26 Apr Shift 1)

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