In a $Δ\ ABC$, $AD$ is the bisector of $∠\ A$, meeting side $BC$ at $D$.
If $BD\ =\ 2\ cm$, $AB\ =\ 5\ cm$, and $DC\ =\ 3\ cm$, find $AC$.
"
Given:
In a $Δ\ ABC$, $AD$ is the bisector of $∠\ A$, meeting side $BC$ at $D$.
$BD\ =\ 2\ cm$, $AB\ =\ 5\ cm$, and $DC\ =\ 4.2\ cm$.
To do:
We have to find the measure of $AC$.
Solution:
$AD$ is the bisector of $∠\ A$, this implies,
$\angle BAD=\angle CAD$
We know that,
The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the corresponding sides containing the angle.
Therefore,
$\frac{AB}{AC} = \frac{BD}{DC}$
$\frac{5}{AC} = \frac{2}{3}$
$AC = \frac{5\times3}{2}$
$AC = 7.5 cm$
The measure of $AC$ is $7.5 cm$. 
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