Question

In $△ABC$, if $AD\perp BC$ and ${AD}^2=BD\times DC$, prove that $\angle BAC={90}^0$.

Answer 1

We have,
${AD}^2=BD\times DC$

$\Rightarrow$ $AD\times AD=BD\times DC$

$\Rightarrow$ $\frac{AD}{DC}=\frac{BD}{AD}$

Thus, in $△ABD$ and $△ACD$, we have

$\frac{AD}{DC}=\frac{BD}{AD}$

and, $\angle BDA=\angle CDA$ [Each equal to ${90}^{\circ }$]

So, by SAS-criterion of similarity, we get

$△DBA\sim △DAC$

$\Rightarrow$ $△DBA$ and $△DAC$ are equiangular

$\Rightarrow$ $\angle 1=\angle C$ and $\angle 2=\angle B$

$\Rightarrow$ $\angle 1+\angle 2=\angle B+\angle C$

$\Rightarrow$ $\angle A=\angle B+\angle C$ [$\because \angle 1+\angle 2=\angle A$]

But, $\angle A+\angle B+\angle C={180}^{\circ }$

$\therefore$ $\angle A+\angle A={180}^0$ [$\because \angle B+\angle C=\angle A$]

$\Rightarrow$ $2\angle A={180}^0\Rightarrow \angle A={90}^{\circ }$

Hence, $\angle BAC={90}^0$