In the given figure,angle BAC = 90^{circ} and | vedclass

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(C) In $	riangle ADB$ and $	riangle ADC$:$\angle ADB = \angle ADC = 90^{\circ}$ (since $AD \perp BC$)$\angle BAD = \angle ACD$ (both are complementa

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$BD \cdot CD = AD^{2}$

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(C) In $	riangle ADB$ and $	riangle ADC$:$\angle ADB = \angle ADC = 90^{\circ}$ (since $AD \perp BC$)$\angle BAD = \angle ACD$ (both are complementary to $\angle CAD$)$\angle ABD = \angle CAD$ (both are complementary to $\angle BAD$)Therefore,by $AA$ similarity criterion,$	riangle ADB \sim 	riangle CDA$.From the property of similar triangles,the ratio of corresponding sides is equal:$\frac{AD}{CD} = \frac{BD}{AD}$Cross-multiplying gives:$AD^{2} = BD \cdot CD$

In the given figure,$\angle BAC = 90^{\circ}$ and $AD \perp BC$. Then,

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