Orthocenter problem: "In an acute-angled ABC H is orthocenter, M is the midpoint of BC...."

[svetli06]

In an acute-angled ABC H is orthocenter, M is the midpoint of BC. The line passing through H and perpendicular to HM intersects sides AC and AB in points B1 and C1 respectively. Prove that H is the midpioint of B1C1.

 

[blamocur]

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[The Highlander]

In an acute-angled ABC H is orthocenter, M is the midpoint of BC. The line passing through H and perpendicular to HM intersects sides AC and AB in points B1 and C1 respectively. Prove that H is the midpioint of B1C1.

I would re-word your question slightly to:-

"In an acute-angled ΔABC, H is the orthocenter and M is the midpoint of BC. The line passing through H and perpendicular to HM intersects sides AC and AB in points B₁ and C₁ respectively. Prove that H is the midpoint of B₁C₁."

And (if that is correct) then a good starting point is always a sketch. Given the one I've made for you (below) what are your thoughts on how to proceed?

​

 

[svetli06]

I have come up with a very complicared solution, so I am looking for easier ones.
Here is my solution very briefly.
Let the ray MH-> intersect (ABC) at K. Thus K lies on (B'HC'A), where B' and C' are the heels of the heights of the triangle. AK||B1C1. AK, B'C', AB are concurrent (can be proven in many ways - pole and polar, Brokar theorem, radical axis). Then we use double ratio for the harmonical quadruple (A1=B'C' intersect sign AB, A' - heel of height from A: B, C) and then use AK||B1C1 to prove that H is the midpoint of B1C1 using a lemma about harmonical quadruples.

 

[Dr.Peterson]

I have come up with a very complicared solution, so I am looking for easier ones.

Thanks for complying with our requests. Due to a peculiarity of the system here, I hadn't seen that you had responded until just now, when I came looking to see if it had been solved yet, having done so myself.

I simply played with the picture, labeling pairs of congruent angles (e.g. the complement of the complement of a vertical angle), and discovered two pairs of similar triangles that, together, nicely prove the theorem.