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Question: If p, q, r are the lengths of the internal bisectors of angles A, B, C of a ∆ABC respectively, then ...

If p, q, r are the lengths of the internal bisectors of angles A, B, C of a ∆ABC respectively, then 1p\frac{1}{p}cosA2\frac{A}{2}+ 1q\frac{1}{q}cosB2\frac{B}{2}+ 1r\frac{1}{r}

cosC2\frac{C}{2}=

A

1a\frac{1}{a}+ 1b\frac{1}{b}–1c\frac{1}{c}

B

1a\frac{1}{a}+1c\frac{1}{c}–1b\frac{1}{b}

C

1a\frac{1}{a}+ 1b\frac{1}{b}+1c\frac{1}{c}

D

1b\frac{1}{b}+1c\frac{1}{c}–1a\frac{1}{a}

Answer

1a\frac{1}{a}+ 1b\frac{1}{b}+1c\frac{1}{c}

Explanation

Solution

∆ =12\frac{1}{2}pc sinA2\frac{A}{2}+ 12\frac{1}{2}pb sin A2\frac{A}{2}

∆ = 12\frac{1}{2}bc sin A = bc sinA2\frac{A}{2} cosA2\frac{A}{2}

= 12\frac{1}{2} pc sinA2\frac{A}{2} + 12\frac{1}{2}pb sinA2\frac{A}{2}

∴ 1p\frac{1}{p}cos A2\frac{A}{2}=12\frac { 1 } { 2 } (1b+1c)\left( \frac{1}{b} + \frac{1}{c} \right)

∴ ∑1pcos⁡A2\sum_{}^{}{\frac{1}{p}\cos\frac{A}{2}} =1a\frac{1}{a}+ 1b\frac{1}{b}+1c\frac{1}{c}