Question

In given figure, If $AD\perp BC$ and $\frac{BD}{DA}=\frac{DA}{DC}$, prove that $△ABC$ is a right triangle.

Answer 1

In ${△}^{\prime }sBDA$ and $ADC$, we have

$\frac{DB}{DA}=\frac{DA}{DC}$ [Given]

$\angle BDA=\angle ADC$ [Each equal to ${90}^{\circ }$]

So, by SAS-criterion of similarity, we have

$△BDA\sim △ADC$

$\Rightarrow$ $\angle ABD=\angle CAD$ and $\angle BAD=\angle ACD$

$\Rightarrow$ $\angle ABD+\angle ACD=\angle CAD+\angle BAD$

$\Rightarrow$ $\angle B+\angle C=\angle A$

$\Rightarrow$ $\angle A+\angle B+\angle C=2\angle A$ [Adding $\angle A$ on both sides]

$\Rightarrow$ $2\angle A={180}^0$

$\Rightarrow$ $\angle A={90}^0$

$\Rightarrow$ $△ABC$ is a right triangle. $\left[Henceproved\right]$