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(A) Let $\alpha, \beta, \gamma$ be the lengths of the tangents from vertices $A, B, C$ to the incircle respectively.Then the sides of the triangle are

Vedclass · Accepted Answer

$\alpha+\beta+\gamma=\frac{1}{r^2}(\alpha \beta \gamma)$

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(A) Let $\alpha, \beta, \gamma$ be the lengths of the tangents from vertices $A, B, C$ to the incircle respectively.Then the sides of the triangle are $a = \beta + \gamma$,$b = \alpha + \gamma$,and $c = \alpha + \beta$.The semi-perimeter $s$ is given by $s = \frac{a+b+c}{2} = \frac{(\beta+\gamma) + (\alpha+\gamma) + (\alpha+\beta)}{2} = \alpha + \beta + \gamma$.The area of the triangle $\Delta$ is given by Heron's formula: $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$.Substituting the values: $\Delta = \sqrt{(\alpha+\beta+\gamma)(\alpha)(\beta)(\gamma)} = \sqrt{s \alpha \beta \gamma}$.We know that the inradius $r$ is given by $r = \frac{\Delta}{s}$.Therefore,$r^2 = \frac{\Delta^2}{s^2} = \frac{s \alpha \beta \gamma}{s^2} = \frac{\alpha \beta \gamma}{s}$.Substituting $s = \alpha + \beta + \gamma$,we get $r^2 = \frac{\alpha \beta \gamma}{\alpha + \beta + \gamma}$.Thus,$\alpha + \beta + \gamma = \frac{\alpha \beta \gamma}{r^2}$.

List-$I$	List-$II$
$(A)$ $rr_2 = r_1r_3$	$(I)$ $\angle A = 90^{\circ}$
$(B)$ $r_1 + r_2 = r_3 - r$	$(II)$ $b^2 = c^2 + a^2$
$(C)$ $r_1 = r + 2R$	$(III)$ $\angle C = 90^{\circ}$
	$(IV)$ $\angle B = 120^{\circ}$

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

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