In $ \triangle \mathrm{ABC} $. the bisector of $ \angle \mathrm{A} $ intersects $ \mathrm{BC} $ at $ \mathrm{D} $. If $ \mathrm{AB}=8, \mathrm{AC}=10 $ and $ \mathrm{BC}=9 $, find $ \mathrm{BD} $ and $ \mathrm{DC} $.
Given:
In \( \triangle \mathrm{ABC} \). the bisector of \( \angle \mathrm{A} \) intersects \( \mathrm{BC} \) at \( \mathrm{D} \).
\( \mathrm{AB}=8, \mathrm{AC}=10 \) and \( \mathrm{BC}=9 \).
To do:
We have to find \( BD \) and $DC$.
Solution:
We know that,
An angle bisector of an angle of a triangle divides the opposite side in two segments that are proportional to the other two sides of the triangle.
Let $BD=x$, this implies, $DC=BC-BD=9-x$
Therefore,
$\frac{AB}{AC}=\frac{BD}{DC}$
$\frac{8}{10}=\frac{x}{9-x}$
$4(9-x)=5x$
$36-4x=5x$
$5x+4x=36$
$9x=36$
$x=4$
$\Rightarrow BD=x=4\ cm$
$DC=9-4=5\ cm$
Hence, the value of $BD$ is $4\ cm$ and the value of $DC$ is $5\ cm$ .+
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