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

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(B) In $\Delta ABC$,$\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$.Consider $\Delta ABM$ and $\Delta ABC$.In $\Delta ABM$,$\angle AMB

Vedclass · Accepted Answer

$\angle MBC$

Vedclass · Answer

(B) In $\Delta ABC$,$\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$.Consider $\Delta ABM$ and $\Delta ABC$.In $\Delta ABM$,$\angle AMB = 90^{\circ}$. Thus,$\angle BAM + \angle ABM = 90^{\circ}$.In $\Delta ABC$,$\angle B = 90^{\circ}$. Thus,$\angle BAM + \angle C = 90^{\circ}$.Comparing these two equations,we get $\angle ABM = \angle C$ (or $\angle MCB$).Also,in $\Delta BMC$,$\angle BMC = 90^{\circ}$,so $\angle MBC + \angle C = 90^{\circ}$.Since $\angle BAM + \angle C = 90^{\circ}$ and $\angle MBC + \angle C = 90^{\circ}$,it follows that $\angle BAM = \angle MBC$.

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

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