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In a $Δ\ ABC$, $AD$ is the bisector of $∠\ A$, meeting side $BC$ at $D$.
If $AB\ =\ 5.6\ cm$, $BC\ =\ 6\ cm$, and $BD\ =\ 3.2\ cm$, find $AC$.
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Given:


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


$AB\ =\ 5.6\ cm$, $BC\ =\ 6\ cm$, and $BD\ =\ 3.2\ cm$.


To do:


We have to find the measure of $AC$.


Solution:


From the figure,


$BC=BD+DC$


$DC=BC-BD=(6-3.2)\ cm=2.8\ cm$


$AD$ 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,


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


$\frac{5.6}{AC} = \frac{3.2}{2.8}$


$AC = \frac{5.6\times2.8}{3.2}$


$AC = \frac{19.6}{4}\ cm$


$AC=4.9\ cm$

The measure of $AC$ is $4.9\ cm$. 

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Updated on: 10-Oct-2022

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