In $\Delta XYZ$ and $\Delta MNO$,$\frac{XY}{MN} = \frac{XZ}{NO} = \frac{YZ}{MO}$. Then,the correspondence $\ldots$ is a similarity.

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AM = 12$ and $BM = 12$,find $AC$.

In $\Delta ABC$,$D$ is the midpoint of $\overline{BC}$. $\Delta ABC$ and $\Delta BDE$ are equilateral triangles. Then,the ratio of the areas of $\Delta ABC$ and $\Delta BDE$ is.......

In Figure,$CM$ and $RN$ are respectively the medians of $\Delta ABC$ and $\Delta PQR$. If $\Delta ABC \sim \Delta PQR$,prove that:
$(i)$ $\Delta AMC \sim \Delta PNR$
$(ii)$ $\frac{CM}{RN} = \frac{AB}{PQ}$
$(iii)$ $\Delta CMB \sim \Delta RNQ$

If $\Delta ABC \sim \Delta DEF$ for the correspondence $ABC \leftrightarrow DEF$,and given $AB = 8$,$AC = 10$,$DE = 12$,and $EF = 18$,find the perimeter of $\Delta DEF$.

$\Delta PQR \sim \Delta XYZ$ for the correspondence $PQR \leftrightarrow YZX$. If $m \angle P = 2 m \angle Q$ and $m \angle X = 120^\circ$,then $m \angle Y = \dots$ (in $^\circ$)