The ratio that the orthocenter divides an altitude into

Question

In a book I'm reading, it claimed that the orthocenter $(H)$ of any triangle (apart from right) divided an altitude into this ratio: $$\frac{AH}{HD}=\frac{\cos A}{\cos B \cos C}$$ I have tried and failed to find a proof of this property, can someone provide a proof?

Answer 1

For acute-angled triangle we obtain: $$\frac{AH}{HD}=\frac{\frac{AE}{\cos\measuredangle HAE}}{BD\tan\measuredangle HBD}=\frac{\frac{c\cos\alpha}{\sin\gamma}}{c\cos\beta\cot\gamma}=\frac{\cos\alpha}{\cos\beta\cos\gamma}$$