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In $\Delta ABC$, if $AD \perp BC$ and $AD^2 = BD \times DC$, then prove that $\angle BAC = 90^\circ$.
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Correct figure. $AD^2 = BD \times DC \implies \frac{AD}{DC} = \frac{BD}{AD}$. Also, $\angle ADB = \angle ADC. \therefore \Delta DBA \sim \Delta DAC \implies \angle DBA = \angle DAC$ and $\angle BAD = \angle DCA$. Adding both, $\angle DBA + \angle DCA = \angle DAC + \angle BAD \implies \angle BAC = 90^\circ$._a_1.png)