Chứng minh [tex]\Delta ABD \sim \Delta CBF (g.g)[/tex]
[tex]\Rightarrow \frac{AB}{BD}=\frac{BC}{BF}[/tex]
[tex]\Rightarrow BC\cdot BD=BF\cdot AB[/tex]
Lại có [tex]AE\cdot AC= AF\cdot AB[/tex]
[tex]\Rightarrow BD\cdot BC+AE\cdot AC=AB(AF+FB)=AB^{2}[/tex]