The in-circle of △ A BC touches side BC at D . Then difference between B D and C D ( R is circumradius of △ A BC )

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​ ∣ B D − C D ∣ = ​ r ( cot 2 B ​ − cot 2 C ​ ) ​ = r ​ sin 2 B ​ sin 2 C ​ sin ( 2 B − C ​ ) ​ ​ = ​ ​ ​ 4 R sin 2 A ​ sin 2 B − C ​ ​ = ​ 4 R cos 2 B + C ​ ⋅ sin 2 B − C ​ ​ = ∣2 R ( sin B − sin C ) ∣ = ∣ b − c ∣

Mathematics - Properties of Triangle Question with Solution | TestHub

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