In $\Delta ABC$, $AD \perp BC$ and $AD^2 = BD \times CD$. Prove that $\angle BAC = 90^\circ$. A B D C

Step by Step Explanation:
1. Given:
$AD \perp BC$ and $AD^2 = BD \times CD$

Here, we have to find the value of $\angle BAC$.
2. Now, we have: $AD^2 = BD \times CD \implies \dfrac { BD } { AD } = \dfrac { AD } { CD }$

Now, in $\Delta DBA$ and $\Delta DAC$, we have

\begin{aligned} &\angle BDA = \angle ADC = 90^\circ \space \text{and} \space \dfrac { BD } { AD } = \dfrac { AD } { CD } \\ \therefore& \space\space\Delta DBA \sim \Delta DAC \space\space\space\space\space\space\space\space\space\space [\text{By SAS-similarity} ] \\ &\text { As the corresponding angles of similar triangles are equal.} \\ &\text{ So, } \space\space \angle B = \angle 2 \text{ and } \angle 1 = \angle C \\ \therefore& \space\space \angle 1 + \angle 2 = \angle B + \angle C \\ \implies&\space\space \angle A = \angle B + \angle C \space\space\space\space\space\space\space\space\space\space [\text{where, } \angle A = \angle 1 + \angle 2 ]\\ \implies&\space\space 2 \angle A = \angle A + \angle B + \angle C \space\space\space\space\space\space\space\space\space\space [\text{Adding}\space \angle A \space \text{on both sides. }]\\ \implies&\space\space 2 \angle A = \angle A + \angle B + \angle C = 180^\circ \space\space\space\space\space\space\space\space\space\space [\text{Sum of angles of a triangle is of}\space 180^\circ.]\\ \implies&\space\space \angle A = \angle BAC = 90^\circ. \end{aligned}