In $△ A B C,$ prove that $r_{1} + r_{2} + r_{3} - r = 4 R$

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Detailed Solution

We know that, $r = 4 R \sin (\frac{A}{2}) \sin (\frac{B}{2}) \sin (\frac{C}{2})$

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

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

$r_{3} = 4 R \cos (\frac{A}{2}) \cos (\frac{B}{2}) \sin (\frac{C}{2})$

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

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

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

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

$= 4 R \cos \frac{C}{2} (\sin (\frac{π}{2} - \frac{C}{2})) = 4 R \cos \frac{C}{2} \cos \frac{C}{2}$

$= 4 R \cos^{2} \frac{C}{2} \dots \dots (1)$

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

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

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

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

$= 4 R \sin \frac{C}{2} (\cos (\frac{π}{2} - \frac{C}{2}))$

$= 4 R \sin \frac{C}{2} \cdot \sin \frac{C}{2} = 4 R \sin^{2} \frac{C}{2} . . . . . . . . . . . (2)$

Adding $(1) and (2) r_{1} + r_{2} + r_{3} - r$

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

$= 4 R ({Cos}^{2} \frac{C}{2} + \sin^{2} \frac{C}{2}) = 4 R (1) = 4 R$

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