In a $Δ\ ABC$, $AD$ is the bisector of $∠\ A$, meeting side $BC$ at $D$.
If $AB\ =\ 5.6\ cm$, $BC\ =\ 6\ cm$, and $BD\ =\ 3.2\ cm$, find $AC$.
"
Given:
In a $Δ\ ABC$, $AD$ is the bisector of $∠\ A$, meeting side $BC$ at $D$.
$AB\ =\ 5.6\ cm$, $BC\ =\ 6\ cm$, and $BD\ =\ 3.2\ cm$.
To do:
We have to find the measure of $AC$.
Solution:
From the figure,
$BC=BD+DC$
$DC=BC-BD=(6-3.2)\ cm=2.8\ cm$
$AD$ 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,
$\frac{AB}{AC} = \frac{BD}{DC}$
$\frac{5.6}{AC} = \frac{3.2}{2.8}$
$AC = \frac{5.6\times2.8}{3.2}$
$AC = \frac{19.6}{4}\ cm$
$AC=4.9\ cm$
The measure of $AC$ is $4.9\ cm$. 
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