In Delta ABC,m angle A = 90^{circ} and overli | vedclass

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(C) In a right-angled triangle $\Delta ABC$ where $m \angle A = 90^{\circ}$ and $\overline{AD} \perp \overline{BC}$,the triangle is divided into two t

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$BD \cdot DC$

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(C) In a right-angled triangle $\Delta ABC$ where $m \angle A = 90^{\circ}$ and $\overline{AD} \perp \overline{BC}$,the triangle is divided into two triangles $\Delta ABD$ and $\Delta CAD$ which are similar to the original triangle $\Delta ABC$ and to each other.By the property of similarity,$\Delta ABD \sim \Delta CAD$.Therefore,the ratio of their corresponding sides is equal: $\frac{AD}{BD} = \frac{CD}{AD}$.Cross-multiplying gives $AD^{2} = BD \cdot CD$ or $AD^{2} = BD \cdot DC$.

In $\Delta ABC$,$m \angle A = 90^{\circ}$ and $\overline{AD}$ is an altitude to the hypotenuse $BC$. Then,$AD^{2} = \ldots$

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