In a $Δ\ ABC$, $AD$ is the bisector of $∠\ A$, meeting side $BC$ at $D$.
If $BD\ =\ 2.5\ cm$, $AB\ =\ 5\ cm$, and $AC\ =\ 4.2\ cm$, find $DC$.
"
Given:
In a $Δ\ ABC$, $AD$ is the bisector of $∠\ A$, meeting side $BC$ at $D$.
$BD\ =\ 2.5\ cm$, $AB\ =\ 5\ cm$, and $AC\ =\ 4.2\ cm$.
To do:
We have to find the measure of $DC$.
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,
$\frac{AB}{AC} = \frac{BD}{DC}$
$\frac{5}{4.2} = \frac{2.5}{DC}$
$DC = \frac{2.5\times4.2}{5}$
$DC = 2.1 cm$
The measure of $DC$ is $2.1 cm$. 
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