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(C) In $\Delta ABC$,$D$,$E$,and $F$ are the midpoints of $\overline{AB}$,$\overline{BC}$,and $\overline{CA}$ respectively.By the Midpoint Theorem,$EF

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$CAB$

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(C) In $\Delta ABC$,$D$,$E$,and $F$ are the midpoints of $\overline{AB}$,$\overline{BC}$,and $\overline{CA}$ respectively.By the Midpoint Theorem,$EF \parallel AB$ and $EF = \frac{1}{2} AB$,$DF \parallel BC$ and $DF = \frac{1}{2} BC$,and $DE \parallel AC$ and $DE = \frac{1}{2} AC$.Therefore,$\frac{DE}{AC} = \frac{EF}{AB} = \frac{DF}{BC} = \frac{1}{2}$.By the $SSS$ similarity criterion,$\Delta DEF \sim \Delta CAB$ (since $D$ corresponds to $C$,$E$ corresponds to $A$,and $F$ corresponds to $B$).Thus,the correspondence $DEF \leftrightarrow CAB$ is a similarity.

In $\Delta ABC$,$D$,$E$,and $F$ are the midpoints of $\overline{AB}$,$\overline{BC}$,and $\overline{CA}$ respectively. Then the correspondence $DEF \leftrightarrow \dots$ is a similarity.

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