In a $Δ A B C, A D$ is the bisector of $∠ A,$ , meeting side $B C$ at $D$ .
If $B D = 2 c m, A B = 5 c m$ and $D C = 3 c m,$
Find $A C$ .

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Solution

According to angle bisector theorem,
$\frac{A B}{B D} = \frac{A C}{D C}$
$A C = \frac{A B \times D C}{B D} = \frac{5 \times 3}{2} = 7.5 c m$ .

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