Show that ab−r1r2r3=bc−r2r3r1=ca−r3r1r2

ab−r1r2=(2Rsin⁡A)(2Rsin⁡B)−4Rsin⁡A2cos⁡B2cos⁡C24Rcos⁡A2sin⁡B2cos⁡C2=4R2sin⁡Asin⁡B−4R2cos2⁡C22sin⁡A2cos⁡A22sin⁡B2cos⁡B2=4R2sin⁡Asin⁡B−4R2cos2⁡C2sin⁡Asin⁡B=4R2sin⁡Asin⁡B1−cos2⁡C2=4R2sin⁡Asin⁡Bsin2⁡C2 Now ab−r1r2r3=4R22sin⁡A2cos⁡A22sin⁡B2cos⁡B2sin2⁡C24Rcos⁡A2cos⁡B2sin⁡C2=4Rsin⁡A2sin⁡B2sin⁡C2=rSimilarly, we can show bc−r1r2r1=ca−r2r1r2=r∴ab−r1r2r3=bc−r2r3r1=ca−r3r1r2=r

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Show that $\frac{a b - r_{1} r_{2}}{r_{3}} = \frac{b c - r_{2} r_{3}}{r_{1}} = \frac{c a - r_{3} r_{1}}{r_{2}}$

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Detailed Solution

$a b - r_{1} r_{2} = (2 R \sin A) (2 R \sin B) -$

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

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

$[2 \sin \frac{A}{2} \cos \frac{A}{2}] [2 \sin \frac{B}{2} \cos \frac{B}{2}]$

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

$= 4 R^{2} \sin A \sin B [1 - \cos^{2} \frac{C}{2}]$

$= 4 R^{2} \sin A \sin B \sin^{2} \frac{C}{2}$

$Now \frac{a b - r_{1} r_{2}}{r_{3}} =$

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