Đáp án:
Xét $\triangle ABC$ có
$\widehat{A}+\widehat{B}+\widehat{C}=180^0$
$\Rightarrow \widehat{B}+\widehat{C}=180^0-\widehat{A}$
mà $\widehat{B}-\widehat{C}=20^0$ (gt)
$\Rightarrow \widehat{B}=\dfrac{180^0-\widehat{A}+20^0}{2}=100-\dfrac{\widehat{A}}{2}$
Do $AD$ là tia phân giác nên $\widehat{BAD}=\widehat{DAC}=\dfrac{\widehat{A}}{2}$
Xét $\triangle ABD$ có:
$\widehat{ADC}=\widehat{B}+\widehat{BAD}$ (tính chất góc ngoài của tam giác)
mà $\widehat{B}=100-\dfrac{\widehat{A}}{2}$, $\widehat{BAD}=\dfrac{\widehat{A}}{2}$
$\Rightarrow \widehat{ADC}=100-\dfrac{\widehat{A}}{2}+\dfrac{\widehat{A}}{2}=100^0$
Ta có: $\widehat{ADC}+\widehat{ADB}=180^0$ (kề bù)
$\Rightarrow \widehat{ADB}=180^0-\widehat{ADC}=180^0-100^0=80^0$
