Giải thích các bước giải:
a.Ta có: $a//b\to\widehat{ADC}+\widehat{DCB}=180^o$
Mà $4\widehat{ADC}=5\widehat{DCB}$
$\to 4\widehat{ADC}+4\widehat{DCB}=9\widehat{DCB}$
$\to 4(\widehat{ADC}+\widehat{DCB})=9\widehat{DCB}$
$\to 4\cdot 180^o=9\widehat{DCB}$
$\to \widehat{DCB}=80^o$
$\to \widehat{ADC}=100^o$
b.Ta có: $DH\perp BC\to DH\perp HC\to \widehat{AHC}=90^o$
Lai có: $\widehat{DCH}=\widehat{DCB}=80^o$
$\to \widehat{HDC}=180^o-\widehat{AHC}-\widehat{DCB}=10^o$