I have to prove that in any $\triangle ABC$ with orthocenter $H$, the product of the length of the segments in which $H$ divides each height is constant.
I have tried getting a closed expresion for H , but the solution was too complicated to work with. I also tried seeing if there was some kind of relationship between angles , but I didn't see any.
Some help would be helpful as I don't have much experience in geometry.
Let Ad, BE, CF be the altitudes. AFDC is cyclic. From which, by power of a point, AH.HD = CH.HF.
Added: This is because, by the converse of "angles in the same segment", $\angle AFC = \angle AFC = 90^0$.
Find another suitable cyclic quadrilateral and repeat the process.