Question

In the given figure, $\angle$ A = $\angle$ CED. Prove that $△$ CAB $\sim$ $△$ CED. Also, find the value of $x$.

Answer 1

Consider $△CAB$ and $△CED$

$\angle CAB=\angle CED$ (given)

$\angle ACB=\angle ECD$ (common angle)

So by $AA$ similarity criteria , $△CAB\sim △CED$

Hence proved.

Also corresponding sides are proportional in similar triangles.

So $\frac{CA}{CE}=\frac{AB}{ED}$

$\frac{CD+DA}{CE}=\frac{AB}{ED}$

$\frac{8+7}{10}=\frac{9}{x}$

$\frac{15}{10}=\frac{9}{x}$

$\frac{3}{2}=\frac{9}{x}$

$x=6cm$