Cho tam giác ABC có

- Hướng dẫn giải

$\Delta ACD$ có $\widehat{D_2}=\widehat{B}+\widehat{A_1}$ (góc ngoài tam giác)

$\Delta ABD$ có $\widehat{D_1}=\widehat{C}+\widehat{A_2}$ (góc ngoài tam giác) mà $\widehat{A_1}=\widehat{A_2}$

nên $\widehat{D_2}-\widehat{D_1}=\widehat{B}-\widehat{C}$

$\Rightarrow \widehat{D_2}-\widehat{D_1}=18°$ mà $\widehat{D_2}+\widehat{D_1}=180°$

nên $\widehat{D_2}=\frac{180°+18°}{2}=99°; \widehat{D_1}=\frac{180°-18°}{2}=81°$