In a $Δ\ ABC$, $AD$ is the bisector of $∠\ A$, meeting side $BC$ at $D$. If $AB\ =\ 10\ cm$, $AC\ =\ 14\ cm$, and $BC\ =\ 6\ cm$, find $BD$ and $DC$. "
Given:
In a $Δ\ ABC$, $AD$ is the bisector of $∠\ A$, meeting side $BC$ at $D$.
$AB\ =\ 10\ cm$, $AC\ =\ 14\ cm$, and $BC\ =\ 6\ cm$.
To do:
We have to find the measure of $BD$ and $DC$.
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,
$\frac{AB}{AC} = \frac{BD}{DC}$
$\frac{10}{14} = \frac{x}{6-x}$
$(6-x)\times10 = x \times 14$
$60-10x=14x$
$14x+10x=60$
$24x=60$
$x=\frac{60}{24}$
$x=2.5\ cm$
Therefore,
$BD = 2.5\ cm$ and $DC=6-2.5\ cm=3.5\ cm$.
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