Tam giác ABC có 

D

- Hướng dẫn giải

Vì $\widehat{HIA}+\text{}\widehat{HFA}={180}^0$ nên tứ giác HFAI nội tiếp.

Suy ra: $\widehat{IHF}+\text{}\widehat{IAF}={180}^0\Rightarrow \widehat{IHF}={180}^0-\text{}\widehat{IAF}={80}^0$

Ta có $\left\{\begin{array}{l}\left(\overrightarrow{HA},\overrightarrow{HB}\right)=\widehat{BHA}\\ \left(\overrightarrow{HB},\overrightarrow{HC}\right)=\widehat{BHC}\\ \left(\overrightarrow{HC},\overrightarrow{HA}\right)=\widehat{CHA}\end{array}\right.$

$\Rightarrow \left(\overrightarrow{HA},\overrightarrow{HB}\right)+\left(\overrightarrow{HB},\overrightarrow{HC}\right)+\left(\overrightarrow{HC},\overrightarrow{HA}\right)=\widehat{BHA}+\widehat{BHC}+\widehat{CHA}$

$=2\widehat{BHC}={2.80}^0={160}^0$

Chọn D.