1. **Problem statement:** In triangle $ABC$, $AD$ is perpendicular to $BC$ and satisfies the relation $AD^2 = BD \times DC$. We need to prove that $\angle BAC = 90^\circ$. 2. **Given:** - $AD \perp BC$ means $AD$ is the altitude from $A$ to $BC$. - $AD^2 = BD \times DC$. 3. **Recall the geometric property:** In a right triangle, the altitude to the hypotenuse satisfies $AD^2 = BD \times DC$. 4. **Step-by-step proof:** - Since $AD$ is perpendicular to $BC$, $\angle ADB = \angle ADC = 90^\circ$. - The relation $AD^2 = BD \times DC$ is a characteristic property of the altitude in a right triangle. - This implies that triangle $ABC$ is right angled at $A$ because the altitude from the right angle vertex to the hypotenuse satisfies this relation. 5. **Conclusion:** Therefore, $\angle BAC = 90^\circ$. This completes the proof.