Lời giải:

a) ΔABC ᔕ ΔA′B′C′ nên ta có:

• \[\frac{{A'B'}} = \frac{{B'C'}} = \frac{{A'C'}}\]

• \[\widehat A = \widehat {A'}\]; \[\widehat B = \widehat {B'}\]; \[\widehat C = \widehat {C'}\].

b) ΔDEF ᔕ ΔD′E′F′ nên ta có:

• \[\widehat {D'} = \widehat D = 78^\circ \]

• \[\widehat {F'} = \widehat F = 180^\circ - \left( {78^\circ + 57^\circ } \right) = 45^\circ \].

Vậy \[\widehat {D'} = 78^\circ \]; \[\widehat {F'} = 45^\circ \].

c) ΔMNP ᔕ ΔM′N′P′ nên ta có

\[\frac{{M'N'}} = \frac{{N'P'}} = \frac{{M'P'}} = \frac{1}{2}\] hay \[\frac = \frac{6} = \frac{{M'P'}} = \frac{1}{2}\].

Do đó \[MN = \frac{2},\;M'P' = 20\].