Trong tam giác ABC, ta có: \(\widehat {ABC} + \widehat {BAC} + \widehat {BCA} = 180^\circ \)

Suy ra \(\widehat {BAC} = 180^\circ - \left( {\widehat {ABC} + \widehat {BCA}} \right) = 180^\circ - \left( {135^\circ + 25^\circ } \right) = 20^\circ \).

Do AB // CD nên \(\widehat {ACD} = \widehat {BAC} = 20^\circ \) (hai góc so le trong).

Trong tam giác ACD, ta có: \(\widehat {ADC} + \widehat {ACD} + \widehat {DAC} = 180^\circ \)

Suy ra \(\widehat {DAC} = 180^\circ - \left( {\widehat {ADC} + \widehat {ACD}} \right) = 180^\circ - \left( {70^\circ - 20^\circ } \right) = 90^\circ \).

Vậy \(\widehat {DAC} = 90^\circ \).