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In $Δ A B C$ , if the internal bisector of $∠ A$ meets $B C$ in $D$ , then

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\frac{A B}{A C} = \frac{B D}{D C}

\frac{A B}{D C} = \frac{A C}{B D}

\frac{A B}{A C} = \frac{D C}{B D}

\frac{A C}{A B} = \frac{B D}{B C}

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detailed solution

Correct option is A

By vertical angular bisector theorem

$\frac{A B}{A C} = \frac{B D}{D C}$

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In ΔABC , if the internal bisector of ∠A meets BC in D , then