In $△ A B C$ , $A B = 3 c m, A C = 4 c m$ and $A D$ is the bisector of $∠ A$ . Then $B D : D C$ is:

9 : 16

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16 : 9

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3 : 4

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4 : 3

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Solution

The correct option is C $3 : 4$
In $△ A B C$ ,
AD bisects $∠ A$
By angle bisector theorem,
$\frac{A B}{A C} = \frac{B D}{D C}$
$\frac{B D}{D C} = \frac{3}{4}$

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