Cho hình vẽ bên. Tính tổng 

- Hướng dẫn giải

Xét $\Delta ABI$ có $\widehat{A}+\widehat{B}=180°-\widehat{AIB}$.

Xét $\Delta CDH$ có $\widehat{C}+\widehat{D}=180°-\widehat{CHD}$.

Xét $\Delta EFK$ có $\widehat{E}+\widehat{F}=180°-\widehat{EKF}$.

Suy ra:

$\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}+\widehat{E}+\widehat{F}=540°-\left(\widehat{AIB}+\widehat{CHD}+\widehat{EKF}\right)$

$=540°-\left(\widehat{KIH}+\widehat{IHK}+\widehat{IKH}\right)=540°-180°=360°$