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

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(B) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$,the triangle is divided into two tri

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$AM$ and $AC$

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(B) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$,the triangle is divided into two triangles $\Delta AMB$ and $\Delta BMC$ which are similar to the original triangle $\Delta ABC$ and to each other.By the property of similarity,$\Delta AMB \sim \Delta ABC$.This implies that the ratio of corresponding sides is equal: $\frac{AM}{AB} = \frac{AB}{AC}$.Cross-multiplying gives $AB^2 = AM \cdot AC$.Therefore,$AB$ is the geometric mean of $AM$ and $AC$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $AC$. Then,$AB$ is the geometric mean of.......

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