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In a $Δ\ ABC$, $AD$ is the bisector of $∠\ A$, meeting side $BC$ at $D$.
If $BD\ =\ 2\ cm$, $AB\ =\ 5\ cm$, and $DC\ =\ 3\ cm$, find $AC$.

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Given:


In a $Δ\ ABC$, $AD$ is the bisector of $∠\ A$, meeting side $BC$ at $D$.


$BD\ =\ 2\ cm$, $AB\ =\ 5\ cm$, and $DC\ =\ 4.2\ cm$.


To do:


We have to find the measure of $AC$.


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,


$\frac{AB}{AC} = \frac{BD}{DC}$


$\frac{5}{AC} = \frac{2}{3}$


$AC = \frac{5\times3}{2}$


$AC = 7.5 cm$

The measure of $AC$ is $7.5 cm$. 

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Updated on: 10-Oct-2022

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