If BE and CF are altitudes of the acute-angled triangle ABC, prove that AF · AB = AE · AC

Question

If BE and CF are altitudes of the acute-angled triangle ABC, prove that AF · AB = AE · AC

Answer 1

Hint:

• Notice that the triangles $ABE$ and $ACF$ are similar, thus...
• or notice that $BCEF$ is a cyclic quadrilateral and thus...

Answer 2

We have $$\cos(\alpha)=\frac{AF}{b}$$ and $$\cos(\alpha)=\frac{AE}{c}$$