Giải thích các bước giải:
a) Đặt \(\alpha = \widehat{B} - \widehat{C} =40^0\).
Ta có: \(\widehat{ADC} =180^0 -( \widehat{C} + \widehat{CAD})\) và
\(\widehat{ADB} =180^0 -( \widehat{B} + \widehat{BAD})\).
Suy ra \(\widehat{ADC}-\widehat{ADB}= \widehat{B}+\widehat{BAD}-(\widehat{C}+\widehat{CAD})\)\(=(\widehat{B}+\frac{\widehat{BAC}}{2})-(\widehat{C}+\frac{\widehat{BAC}}{2})=\widehat{B}-\widehat{C}\)
\(\widehat{ADC}-\widehat{ADB}=\alpha \) (1). Dễ dàng suy ra \(\widehat{ADC}>90^0>\widehat{ADB}\)
\(\Rightarrow H\) nằm giữa \(D\) và \(B\).
Mặt khác \(\widehat{ADC}+\widehat{ADB}=180^{\circ}\) (2)
Từ (1) và (2) suy ra \(\widehat{ADC}=90^{\circ}+\frac{\alpha }{2}; \widehat{ADB}=90^{\circ}-\frac{\alpha }{2}\)
b) Trong tam giác HAD, ta có: \(\widehat{HAD} + \widehat{AHD} + \widehat{ADH} = 180^0 \)
\(\Rightarrow \widehat{HAD} = 180^0 - \widehat{AHD} - \widehat{ADH}=180^0 - 90^{\circ}-\widehat{ADH}=90^{\circ}-(90^{\circ}-\frac{\alpha }{2})=\frac{\alpha }{2}\).
Vậy \(\widehat{HAD} = \frac{\alpha }{2} = \frac{40^0}{2} = 20^0\).
