In $\Delta ABC$,$m\angle B = 90^\circ$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. Prove the following:
$(i)$ The correspondence $AMB \leftrightarrow ABC$ is a similarity.
(ii) The correspondence $BMC \leftrightarrow ABC$ is a similarity.
(iii) The correspondence $AMB \leftrightarrow BMC$ is a similarity.

(A) In $\Delta ABC$,$m\angle B = 90^\circ$ and $\overline{BM} \perp \overline{AC}$.
$(i)$ For the correspondence $AMB \leftrightarrow ABC$:
$\angle AMB \cong \angle ABC$ (both are $90^\circ$)
$\angle MAB \cong \angle BAC$ (common angle)
Therefore,by the $AA$ similarity criterion,$\Delta AMB \sim \Delta ABC$.
(ii) For the correspondence $BMC \leftrightarrow ABC$:
$\angle BMC \cong \angle ABC$ (both are $90^\circ$)
$\angle MCB \cong \angle BCA$ (common angle)
Therefore,by the $AA$ similarity criterion,$\Delta BMC \sim \Delta ABC$.
(iii) Since $\Delta AMB \sim \Delta ABC$ and $\Delta BMC \sim \Delta ABC$,by the transitivity of similarity,$\Delta AMB \sim \Delta BMC$.