bn tự vẽ hình nha.
ta có: \(\widehat{B}+\widehat{C}=90^0\)
\(\widehat{FBC}+\widehat{FCB}=\dfrac{2}{3}\left(\widehat{ABC}+\widehat{ACB}\right)\)
\(=60^0\)
Xét tam giác BFC có:
\(\widehat{BFC}=180^0-\left(\widehat{FBC}+\widehat{FCB}\right)\)
\(=180^0-60^0=120^0\left(đpcm\right)\)