Cho hình vẽ sau, trong đó

- Hướng dẫn giải

Kẻ CK // AB .Ta có $\text{AB}//\text{ CK}\Rightarrow {\widehat{C}}_1=\widehat{B}={40}^0$ ( hai góc so le trong )

$\left.\begin{array}{l}\text{CK\hspace{0.17em}//\hspace{0.17em}AB}\\ \text{DE\hspace{0.17em}//\hspace{0.17em}AB}\end{array}\right\}\Rightarrow \text{\hspace{0.17em}\hspace{0.17em}CK\hspace{0.17em}//\hspace{0.17em}DE}$

$\text{CK}//\text{ DE}\Rightarrow {\widehat{C}}_2=\widehat{E}={30}^0$ ( hai góc so le trong )

Do đó $\widehat{C_1}+\widehat{C_2}={40}^0+{30}^0={70}^0\Rightarrow \widehat{BCE}={70}^0$