Question

In triangle $ABC,AD$ is perpendicular to $BC$ and $A{D}^2=BD\times DC$. Prove that angle $BAC={90}^0$.

Answer 1

{ Note:signs in figure show propotional sides ;not congruent sides}
$A{D}^2=BD\times DC$
$\Rightarrow AD\times AD=BD\times DC$
$\Rightarrow \frac{AD}{BD}=\frac{DC}{AD}$
In $\Delta ABDand\Delta ACD$
$\frac{AD}{BD}=\frac{DC}{AD}$ and $\angle ADB=\angle ADC={90}^0$
$\Rightarrow \Delta ABD\sim \Delta ACD$

{ Note:signs in figure shows propotional sides }
$\Rightarrow \angle x=\angle b$ and $\angle y=\angle a$
[In similar $\Delta$s, corresponding angles are equal]
In right angle $\Delta ABD$,
$\angle x+\angle y={90}^0\Rightarrow \angle x+\angle a=90[\therefore \angle y=\angle a]\Rightarrow \angle BAC={90}^0$
Hence proved.