Tam giác ABC có góc B > góc C Tia phân giác góc BAC

* Hướng dẫn giải

a) $\Delta ACD$ có $\widehat{A_2}+\widehat{C}+\widehat{ADC}=180°$;

$\Delta ABD$ có $\widehat{A_1}+\widehat{B}+\widehat{ADB}=180°$;

Mà $\widehat{A_1}=\widehat{A_2}$ nên $\widehat{C}+\widehat{ADC}=\widehat{B}+\widehat{ADB}\Rightarrow \widehat{ADC}-\widehat{ADB}=\widehat{B}-\widehat{C}\sqrt{a^2+b^2}$.

b) $\Delta ABC$ có $\widehat{BAx}=\widehat{B}+\widehat{C}$ (góc ngoài tam giác)

$\Rightarrow \widehat{A_3}=\widehat{A_4}=\frac{1}{2}\widehat{BAX}=\frac{\widehat{B}+\widehat{C}}{2}$

$\Delta ACE$ có: $\widehat{A_4}=\widehat{E}+\widehat{C}$ (góc ngoài)

$\Rightarrow \widehat{E}=\widehat{A_4}-\widehat{C}\Rightarrow \widehat{AEB}=\frac{\widehat{B}+\widehat{C}}{2}-\widehat{C}$ hay $\widehat{AEB}=\frac{\widehat{B}-\widehat{C}}{2}$