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 $DC$.
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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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Updated on: 10-Oct-2022

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