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?
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For acute-angled triangle we obtain: $$\frac{AH}{HD}=\frac{\frac{AE}{\cos\measuredangle HAE}}{BD\tan\measuredangle HBD}=\frac{\frac{c\cos\alpha}{\sin\gamma}}{c\cos\beta\cot\gamma}=\frac{\cos\alpha}{\cos\beta\cos\gamma}$$
answered Jun 23, 2018 at 2:36
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1$\begingroup$ I'm assuming this works for obtuse-angled triangles if $AH$ and $HD$ are directed, because one of $\cos\alpha , \cos\beta ,\cos\gamma$ is negative. $\endgroup$ Jun 23, 2018 at 3:13