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In a $Δ\ ABC$, $AD$ is the bisector of $∠\ A$, meeting side $BC$ at $D$.
If $AC\ =\ 4.2\ cm$, $DC\ =\ 6\ cm$, and $BC\ =\ 10\ cm$, find $AB$.
"
Given:
In a $Δ\ ABC$, $AD$ is the bisector of $∠\ A$, meeting side $BC$ at $D$.
$AC\ =\ 4.2\ cm$, $DC\ =\ 6\ cm$, and $BC\ =\ 10\ cm$.
To do:
We have to find the measure of $AB$.
Solution:
From the figure,
$BC=BD+DC$
$BD=BC-DC=(10-6)\ cm=4\ cm$
$AD$ 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,
$\frac{AB}{AC} = \frac{BD}{DC}$
$\frac{AB}{4.2} = \frac{4}{6}$
$AB = \frac{4.2\times4}{6}$
$AB = \frac{8.4}{3}\ cm$
$AB=2.8\ cm$
The measure of $AB$ is $2.8\ cm$. 
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