In a ∆ABC, it is given that AB = 6 cm, AC = 8 cm and AD is the bisector of ∠A.


In a ∆ABC, it is given that AB = 6 cm, AC = 8 cm and AD is the bisector of ∠A. Then, BD DC =?

(a) 3 : 4
(b) 9 : 16
(c) 4 : 3

(d) $\sqrt{3}: 2$



(a) 3 : 4

In $\triangle A B D$ and $\triangle A C D$, we have:

$\angle B A D=\angle C A D$


$\frac{B D}{D C}=\frac{A B}{A C}=\frac{6}{8}=\frac{3}{4}$

$B D: D C=3: 4$


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