In a triangle $A B C$ , with usual notations the length of the bisector of internal angle $A$ is?

\frac{2 b c cos \frac{A}{2}}{b + c}

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\frac{2 b c sin \frac{A}{2}}{b + c}

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\frac{a b c cosec \frac{A}{2}}{2 R (b + c)}

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\frac{2 Δ}{b + c} \cdot cosec \frac{A}{2}

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Δ A B C

b^{2} = c^{2} + a^{2} - 2 c a cos B

A B C

{(a - b)}^{2} {cos}^{2} \frac{C}{2} + {(a + b)}^{2} {sin}^{2} \frac{C}{2} = c^{2}

△ A B C

(a - b)^{2} {cos}^{2} (\frac{C}{2}) + (a + b)^{2} {sin}^{2} (\frac{C}{2}) = c^{2}

A B C

2 a cos (\frac{B - C}{2}) = b + c

B \neq C

A

With usual notations, is a $△ A B C \frac{b^{2} - c^{2}}{a s e c c} + \frac{c^{2} - a^{2}}{b s e c c} + \frac{a^{2} - b^{2}}{c s e c c}$ is equal to

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In a triangle ABC, with usual notations the length of the bisector of internal angle A is?