Let M, N be midpionts of AB, AC and (O) circumcircle of $\triangle ABC.$ Let the A-altitude $AP \perp MN$ meet MN at X, BC at D and (O) again at Q. Then DP = DQ. Denote h = DA, p = XP, m = XM, n = XN. Power of D to (O) is $4 mn = DB \cdot DC = DA \cdot DQ = h(h/2+p).$ This quadratic equation for h or x = h/2 is easy to solve: $x^{2}+px-2mn = 0,$ $x =-p/2 \pm \sqrt{p^{2}/4+2mn}.$ Let (K) be circle with diameter MN, Y one intersection of (K) with XP and XYZU a square. Then $XZ = \sqrt{2mn}.$ Circle (X) with this radius meets MN at Z'. Let X' be midpoint of XP. Then $X'Z' = \sqrt{p^{2}/4+2mn}.$ Circle (X') with this radius meets XP at D, D', the 2 possible A-altitude feet. DB = D'B' = 2XM, DC = D'C' = 2XN on parallels to MN through D, D'. BM meet CN at A on XP and B'M meet C'N at A' also on XP.

Jutaro wrote: Let $ ABC$ a triangle, $ M$, $ N$ the midpoints of $ AB,AC$ and $ H$ the orthocenter. Given the points $ M,N,H$ construct the triangle If the triangle $ MHN$ is right then $ \triangle ABC$ is also right, which means that $ H$ coincides with some vertice (if it is right at $ H$ then $ H\equiv A$, if it is right at $ M$ then $ H\equiv B$, if it is right at $ N$ then $ H\equiv C$) So suppose that $ \triangle MHN$ is not right Let $ w$ the circumcircle of $ \triangle ABC$. Let $ F$ be the anti-diametric point of $ B$. It is $ FA \perp BA \Rightarrow FA \parallel HC$ and $ FC \perp BC \Rightarrow FC\parallel AH$. (notice that noone of the above segments is zero, because $ \triangle ABC$ is not right) So $ AHCF$ is a parallelogram. This means that $ N$ is the midpoint of $ FH$ So the points $ H,F$ are symmetric wrt $ N$ Construction Let $ \varepsilon$ be the line peprendicular to $ MN$ passing through $ H$. Clearly $ A\in \varepsilon$ Let $ F$ be the symmetric point of $ H$ wrt $ N$. The circle with diameter $ MF$ intersects the line $ \varepsilon$ at two points $ A_{1},A_{2}$ For $ A\in\{A_{1},A_{2}\}$ let $ B,C$ be the symmetric points of $ A$ wrt $ M,N$ respectively Proof From the construction it is $ AF\perp AB$. But $ AHCF$ is a parallelogram (because $ N$ is the midpoint of $ AC$ and $ HF$) So $ CH\perp AB$. On the other hand it is $ AH\perp MN \Rightarrow AH\perp BC$, hence $ H$ is the orthocenter of $ \triangle ABC$