Ta cần chứng minh \(\overrightarrow{BF}.\overrightarrow{FG}=0\)
Ta có \(\overrightarrow{BF}=\frac{1}{2}\left(\overrightarrow{BA}+\overrightarrow{BE}\right)\)
\(\overrightarrow{FG}=\frac{1}{2}\left(\overrightarrow{FD}+\overrightarrow{FC}\right)=\frac{1}{2}\left(\overrightarrow{FA}+\overrightarrow{AD}+\overrightarrow{FE}+\overrightarrow{EC}\right)=\frac{1}{2}\left(\overrightarrow{AD}+\overrightarrow{EC}\right)\)
=> \(\overrightarrow{BF}.\overrightarrow{FG}=\frac{1}{4}\left(\overrightarrow{BA}+\overrightarrow{BE}\right)\left(\overrightarrow{AD}+\overrightarrow{EC}\right)=\frac{1}{4}\left(\overrightarrow{BA}.\overrightarrow{AD}+\overrightarrow{BA}.\overrightarrow{EC}+\overrightarrow{BE}.\overrightarrow{AD}+\overrightarrow{BE}.\overrightarrow{EC}\right)\)
\(=\frac{1}{4}\left(0+\overrightarrow{BA}.\overrightarrow{EC}+\overrightarrow{BE}.\overrightarrow{AD}+0\right)=\frac{1}{4}\left(\overrightarrow{BA}.\overrightarrow{EC}+\overrightarrow{BE}.\overrightarrow{AD}\right)\)
\(=\frac{1}{4}\left(\overrightarrow{EA}.\overrightarrow{EC}+\overrightarrow{BE}.\overrightarrow{BC}\right)\) (vì EA là hình chiếu của BA lên EC; AD song song và bằng BC)
\(=\frac{1}{4}\left(-BE^2+\overrightarrow{BE}.\overrightarrow{BC}\right)\) (tính chất đường cao tam giác vuông BAC)
\(=\frac{1}{4}\overrightarrow{BE}\left(-\overrightarrow{BE}+\overrightarrow{BC}\right)=\frac{1}{4}\overrightarrow{BE}\left(\overrightarrow{EB}+\overrightarrow{BC}\right)=\frac{1}{4}\overrightarrow{BE}.\overrightarrow{EC}=0\)
(ĐFCM)
