Kẻ DH,BK vuông góc với AC
Dễ thấy:
\(\Delta\)ADH ~ \(\Delta\)AOF ( g.g ) \(\Rightarrow\frac{AD}{AO}=\frac{DH}{OF}=\frac{FA}{AH}\)
\(\Delta\)ABK ~ \(\Delta\)AOE ( g.g ) \(\Rightarrow\frac{AB}{AO}=\frac{BK}{OE}=\frac{KA}{EA}\)
\(\Rightarrow\frac{BK}{OE}\cdot\frac{OF}{DH}=\frac{AB}{AO}\cdot\frac{AO}{AD}\Rightarrow\frac{OF}{OE}=\frac{AB}{AD}\left(đpcm\right)\)