In $\displaystyle \triangle \mathrm{ABC}$, if $\displaystyle \mathrm{AD} \perp \mathrm{BC}$ and $\displaystyle \mathrm{AD}^{2}=\mathrm{BD} \times \mathrm{DC}$, then prove that $\displaystyle \angle \mathrm{BAC}=90^{\circ}$. /CD1BA/$\displaystyle 22$ This section consists of $\displaystyle 3$ case study based questions of $\displaystyle 4$ marks each.

\[\begin{gathered} \mathrm{AD}^{2}=\mathrm{BD} \times \mathrm{DC} \\ \frac{\mathrm{AD}}{\mathrm{DC}}=\frac{\mathrm{BD}}{\mathrm{AD}} \end{gathered} \] \[\begin{gathered} \triangle \mathrm{ADB} \sim \triangle \mathrm{CDA} \\ \angle \mathrm{BAD}=\angle \mathrm{ACD} \& \angle \mathrm{DAC}=\angle \mathrm{DBA} \\ \text { In } \triangle \mathrm{ABC} \\ \angle \mathrm{ABC}+\angle \mathrm{CAB}+\angle \mathrm{BCA}=180^{\circ} \\ \angle \mathrm{ABC}+\angle \mathrm{BAD}+\angle \mathrm{CAD}+\angle \mathrm{BCA}=180^{\circ} \\ \angle \mathrm{BAD}+\angle \mathrm{CAD}=90^{\circ} \\ \angle \mathrm{BAC}=90^{\circ} \end{gathered} \]