Question

In given figure, $AB\perp BC$ and $DE\perp AC$. Prove that $△ABC\sim △AED$.

Answer 1

In ${△}^{\prime }sABC$ and $AED$, we have

$\angle ABC=\angle AED={90}^0$

$\angle BAC=\angle EAD$ [Each equal to $\angle A$]

Therefore, by AA-criterion of similarity, we have

$△ABC\sim △AED$. $\left[Henceproved\right]$