Question

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

• A. ${90}^{\circ }$
• B. ${60}^{\circ }$
• C. ${30}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is A ${90}^{\circ }$

Given: $△ABC$, $A{D}^2=BD\times DC$, $AD\perp BC$
Using Pythagoras Theorem
In $\Delta$ ABD
$A{B}^2=A{D}^2+B{D}^2$
$A{B}^2=BD\times DC+B{D}^2$ $[GivenA{D}^2=BD\times DC]$
$A{B}^2=BD\times (DC+BD)$
$A{B}^2=BD\times BC$ .......... (i)
In $\Delta$ ADC,
$A{C}^2=A{D}^2+D{C}^2$
$A{C}^2=BD\times DC+D{C}^2=DC\times (BD+DC)$
$A{C}^2=DC\times BC$ ..........(ii)
$\therefore A{B}^2+A{C}^2=BD\times BC+DC\times BC$ [Adding (i) and (ii)]
$A{B}^2+A{C}^2=BC\times (BD+DC)$
$A{B}^2+A{C}^2=BC\times BC=B{C}^2$
$\Rightarrow \angle BAC={90}^{\circ }$ [Using converse of Pythagoras Theorem]