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

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Question

(A) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$,if $\overline{BM}$ is the altitude to the hypotenuse $\overline{AC}$,then $\

Vedclass · Accepted Answer

$AM$ and $CM$

Vedclass · Answer

(A) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$,if $\overline{BM}$ is the altitude to the hypotenuse $\overline{AC}$,then $\Delta AMB \sim \Delta BMC$.From the property of similar triangles,the ratio of corresponding sides is equal:$\frac{AM}{BM} = \frac{BM}{CM}$.Cross-multiplying gives $BM^2 = AM \cdot CM$.Therefore,$BM = \sqrt{AM \cdot CM}$,which means $BM$ is the geometric mean of $AM$ and $CM$.

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

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