Giải thích các bước giải:
a.Ta có $\widehat{ADB}=\widehat{DAC}+\widehat{ACB}$
$\to \widehat{ADB}=\dfrac 12\widehat{BAC}+\widehat{ACB}=\dfrac 12(180^o-\widehat{ABC}-\widehat{ACB})+\widehat{ACB}$
$\to \widehat{ADB}=90^o+\dfrac 12(-\widehat{ABC}+\widehat{ACB})$
$\to \widehat{ADB}=90^o-\dfrac 12(\widehat{ABC}-\widehat{ACB})$
$\to \widehat{ADB}=90^o-\dfrac 12.\delta $
$\to \widehat{ADC}=180^o-\widehat{ADC}=90^o+\dfrac 12\delta $
b.Ta có $\widehat{HAD}=90^o-\widehat{ADH}=90^o-\widehat{ADB}=\dfrac 12.\delta $
