The question is: Prove that connecting the feet of the altitudes of a given triangle, we obtain another triangle for with the altitudes of the given triangle are angle bisectors.
I've tried using algebra and using triangle properties, but somehow nothing fits. Please help, math experts!
Let $AA'$, $BB'$, $CC'$ be the heights of acute triangle and $H$ their intersection (the orthocenter).
Consider $AB'HC'$ -- it's concyclic (inscribed) as $\angle B' = \angle C' = 90^\circ$, hence $\angle HB'C' = \angle HAC'$. Analogically, $\angle HB'A' = \angle HCA'$.
Hence $$\angle HB'A' = \angle HCA' = \angle C'CB = 90^\circ - \angle C'BC = 90^\circ - \angle ABC,$$ $$\angle HB'C' = \angle HAC' = \angle A'AB = 90^\circ - \angle A'BA = 90^\circ - \angle CBA,$$ so these two are equal.
For obtuse triangle the same argument might be applied, with minor changes.