First we proof that $ANA_1M$ lie on the same circle. Observe that $A'NCA_1$ are concyclic, since $\angle NA'A_1=\angle NCA_1=90^o$, from here $\angle NA_1A'=\angle A'CN=\angle ACB$. Now it is easy to see that $\angle ANA'=\angle ABC$, hence $\angle AMN=180-(\angle BAC+\angle ABC)=\angle ACB=\angle NA_1A'$ so they lie in the same circle. Now, in order to proof that the center of the circle lie on the height from $A$ of the triangle $ABC$ we draw it, and call $P$ the intersection of it to the circle $AMA_1N$. Lets evaluate the angle $\angle AA_1P$. Since $PANA_1$ li on the same circle $\angle NA_1P=180-\angle NAP=180-(90-\angle ACB)=90+\angle ACB$. Now $\angle AA_1P=\angle NA_1P-\angle NA_1A'=90+\angle ACB-\angle ACB=90^o$ so $APA_1$ is a right-angle triangle whose circumcircle is the same as $AMA_1N$ hence the center lies on $AP$ which we said to be the height from $A$. A pretty easy problem.

Attachments:

∠$ANM$ = 90 - ∠$CAA'$ = ∠$ABC$ => $MBNC$ is a cyclic quadrilateral => $A'M . A'N = A'B . A'C = A'A1 . A'A $ => $AMA_1N$ is a cyclic quadrilateral $O_1$ is the center of $(AMA_1N)$ => ∠$MAO_1$= ∠$NAA'$ => $AO_1$ ⊥ $BC$

$AC$ is a diameter in $(ABC)$ so $\angle ACA_1 = 90^{\circ} = \angle AA’N$. $\Rightarrow A’NCA_1$ is cyclic. $\Rightarrow \angle A’NA_1 \cong A’CA_1 \cong^{ ABA_1C cyclic} MAA_1 \Rightarrow MANA_1 cyclic $. Let $O’$ be the circumcenter of triangle $\triangle ABC$.$MA’$ altitude so it is isogonal to $MO’$. $\Rightarrow MA’O \cong AMO’ \cong NMA_1 \cong A_1AN $.$\Rightarrow AA’$ is the diameter in $\triangle AMN$ and it is isogonal to $A’O \Rightarrow $ it is the altitude q.e.d.

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%204.pdf p. 22-24. Sincerely Jean-Louis