In $△ A B C If r + r_{1} + r_{2} = r_{3}$ Then $show that ∠ C = 90^{\circ}$

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Detailed Solution

$r_{1} + r_{2} = r - r_{3} \Rightarrow \frac{r_{1} + r_{2}}{r - r_{3}} = 1 . . . . (1)$

$r_{1} + r_{2} = 4 R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} + 4 R$

$\cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}$

$= 4 R \cos \frac{C}{2} [\sin \frac{A}{2} \cos \frac{B}{2} + \cos \frac{A}{2} \sin \frac{B}{2}]$

$= 4 R \cos \frac{C}{2} [\sin \frac{(A + B)}{2}]$

$= 4 R \cos \frac{C}{2} [\cos \frac{C}{2}]$

$= 4 R \cos^{2} \frac{C}{2}$

$r - r_{3} = 4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$

$- 4 R \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}$

$= 4 R \sin \frac{C}{2} [\sin \frac{A}{2} \sin \frac{B}{2} - \cos \frac{A}{2} \cos \frac{B}{2}]$

$= 4 R \sin \frac{C}{2} [- \cos (\frac{A + B}{2})]$

$= 4 R \sin \frac{C}{2} [- \sin \frac{C}{2}] = 4 R \sin^{2} \frac{C}{2}$

$\Rightarrow \frac{r - r_{3}}{r_{1} + r_{2}} = \frac{4 R \sin^{2} \frac{C}{2}}{4 R \cos^{2} \frac{C}{2}} = {Tan}^{2} \frac{C}{2}$

$∴ {Tan}^{2} \frac{C}{2} = Tan 45^{\circ}$

$\frac{C}{2} = 45^{0} ∴ C = 90^{\circ}$

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