Giải thích các bước giải:

a.$\widehat{DAB}, \widehat{ADC}$ là hai góc trong cùng phía

b.$\widehat{DAC}, \widehat{BCA}$ so le trong

c.Ta có: $AB\perp AD, AB\perp BC$

$\to AD//BC$

d.Vì $AD//BC$

$\to \widehat{ADC}+\widehat{DCB}=180^o$

Ta có: $2\widehat{ADC}=3\widehat{DCB}$

$\to \dfrac{\widehat{ADC}}3=\dfrac{\widehat{DCB}}2=\dfrac{\widehat{ADC}+\widehat{DCB}}{3+2}=\dfrac{180^o}5=36^o$

$\to \widehat{ADC}=108^o, \widehat{DCB}=72^o$