Question

In a right angled triangle ABC in which ∠A = ${90}^{\circ }$, If AD $\perp$ BC.

• A. ${AB}^2$ = BD × DC.
• B. ${AB}^2$ = BD × AD.
• C. ${AB}^2$ = BC × DC.
• D. ${AB}^2$ = BC × BD.

Answer 1

The correct option is D

${AB}^2$ = BC × BD.

Consider $\Delta$BAC and $\Delta$BDA,

$\angle$ABC = $\angle$ DBA

$\angle$BAC = $\angle$BDA = ${90}^{\circ }$

$\angle$BCA = $\angle$BAD

$Therefore$, $△ABC$ is similar to $△DBA$ [by AAA]

$\frac{AB}{BD}$ = $\frac{BC}{AB}$. [Sides opposite to equal angles are proportional]

Hence, ${AB}^2$ = $BC\times BD$.