Question

In given figure, $\angle BAC={90}^0$ and segment $AD\perp BC$. Prove that ${AD}^2=BD\times DC$.

Answer 1

In $△ABD$ and $△ACD$, we have

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

$\angle DBA=\angle DAC$ [Each equal to complement of $\angle BAD$ i.e. ${90}^0-\angle BAD$]

Therefore, by AA-criterion of similarity, we have

$△DBA\sim △DAC$ [$\therefore \angle D\leftrightarrow \angle D,\angle B\leftrightarrow \angle DAC$ and $\angle BAD\leftrightarrow \angle DCA$]

$\Rightarrow$ $\frac{DB}{DA}=\frac{DA}{DC}$ [In similar triangles corresponding sides are proportional]

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

$\Rightarrow$ ${AD}^2=BD\times DC$