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

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(A) In $\Delta ADB$ and $\Delta BDC$:$\angle ADB = \angle BDC = 90^{\circ}$ (since $\overline{BD} \perp \overline{AC}$).In $\Delta ABC$,$\angle A + \a

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$BDC$

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(A) In $\Delta ADB$ and $\Delta BDC$:$\angle ADB = \angle BDC = 90^{\circ}$ (since $\overline{BD} \perp \overline{AC}$).In $\Delta ABC$,$\angle A + \angle C = 90^{\circ}$.In $\Delta ADB$,$\angle A + \angle ABD = 90^{\circ}$.Therefore,$\angle ABD = \angle C$.By the $AA$ similarity criterion,the correspondence $A \leftrightarrow B$,$D \leftrightarrow D$,and $B \leftrightarrow C$ holds.Thus,$\Delta ADB \sim \Delta BDC$ is incorrect; the correct correspondence is $\Delta ADB \sim \Delta BDC$ is not the standard notation,but based on the vertices,$\Delta ADB \sim \Delta BDC$ implies $A \leftrightarrow B, D \leftrightarrow D, B \leftrightarrow C$. Looking at the options,the correspondence $ADB \leftrightarrow BDC$ is the correct similarity.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BD}$ is an altitude to the hypotenuse $\overline{AC}$. Then the correspondence $ADB \leftrightarrow \ldots$ between $\Delta ADB$ and $\Delta BDC$ is a similarity.

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