In $ riangle ABC$, if $AD \perp BC$ and $AD^2 = BD imes DC$, then prove that $\angle BAC = 90°$.

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Question

In $	riangle ABC$, if $AD \perp BC$ and $AD^2 = BD 	imes DC$, then prove that $\angle BAC = 90°$.

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Accepted Answer

Step 1: Triangle ABC with altitude AD perpendicular to BC, D on BC
Step 2: $AD^2 = BD 	imes DC \Rightarrow \frac{AD}{BD} = \frac{DC}{AD}$. Also, $\angle ADB = \angle ADC = 90°$.
Step 3: $	herefore 	riangle DBA \sim 	riangle DAC$ (SAS similarity). So $\angle DBA = \angle DAC$ and $\angle BAD = \angle DCA$.
Step 4: Adding both: $\angle DBA + \angle BAD = \angle DAC + \angle DCA$
Step 5: In $	riangle ABD$: $\angle DBA + \angle BAD = 90°$ and in $	riangle ACD$: $\angle DAC + \angle DCA = 90°$. Therefore $\angle BAC = \angle BAD + \angle DAC = 90°$.

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