In a $Δ A B C, A D$ is the bisector of $∠ A,$ meeting side $B C$ at $D$ .
If $A B = 10 c m$ , $A C = 14 c m$ and $B C = 6 c m$ ,
Find $B D$ and $D C$ .

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Solution

According to angle bisector theorem,
$\frac{A B}{B D} = \frac{A C}{D C}$
$B D = \frac{A B \times C D}{A C}$
$B D = \frac{A B \times (6 - B D)}{A C} = \frac{10 \times (6 - B D)}{14}$
$14 B D = 60 - 10 B D$
$B D = 2.5 c m$
$D C = 6 - B D = 6 - 2.5 = 3.5 c m$

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