Altitudes Concur at a Point: a Second Look
The altitudes of a triangle concur at a point - the orthocenter of the triangle. There are multitudes of proofs, each shedding light of a different hue on the existence of the orthocenter. Collecting these proofs was an enjoyable undertaking, and edifying, too. Not until a few days ago, when I came across another problem, have I realized that, while diligently searching and documenting the proofs, I was negligent in one aspect of problem solving.
Looking back is the last step of G. Polya's problem solving methodology that is often omitted once a satisfactory solution to a problem had been found.
So, the altitudes of a triangle concur at a point. Let's see what we can do with that problem. Starting literally from the end, assume there are three lines that meet at a point; is there a triangle for which the three lines serve as the altitudes? If so, how to construct such a triangle?
The answer to the first question is positive and one possible construction is based on a solution to a related problem:
Given three concurrent lines: α, β, and γ. From point A on on α drop perpendiculars AB to γ and AC to β, with B on β and C on γ. Prove that BC is perpendicular to α
The existence of the orthocenter shows the way of solving this problem. Indeed, in ΔABC, lines β and γ are the altitudes. The point of their intersection is the orthocenter, which belongs to the third altitude (the one through A) as well. But the line through A and the orthocenter is exactly α; and the problem is solved making the construction clear.
When point A is selected on one side of the common point of α, β, and γ all triangle the construction leads to are similar. What if A is chosen on the other side of α?
No. It seems that three concurrent lines uniquely define a shape of the triangle for which they serve as altitudes. Is that clear? Are there any special cases?