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In a $Δ\ ABC$, $AD$ is the bisector of $∠\ A$, meeting side $BC$ at $D$.
If $AB\ =\ 5.6\ cm$, $AC\ =\ 6\ cm$, and $DC\ =\ 3\ cm$, find $BC$."
Given:
In a $Δ\ ABC$, $AD$ is the bisector of $∠\ A$, meeting side $BC$ at $D$.
$AB\ =\ 5.6\ cm$, $AC\ =\ 6\ cm$, and $DC\ =\ 3\ cm$.
To do:
We have to find the measure of $BC$.
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,
$\frac{AB}{AC} = \frac{BD}{DC}$
$\frac{5.6}{6} = \frac{BD}{3}$
$BD = \frac{3\times5.6}{6}$
$BD = \frac{5.6}{2}\ cm$
$BD=2.8\ cm$
From the figure,
$BC=BD+DC$
$BC=(2.8+3)\ cm=5.8\ cm$
The measure of $BC$ is $5.8\ cm$. 
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