It's a common knowledge that three altitudes in a triangle intersect in a point known as the orthocenter of the triangle. So let's start with a triangle ABC and draw all three of its altitudes AH_a, BH_b, and CH_c.

Now there is an interesting question: How many triangles with altitudes drawn are there in the diagram?

The answer is 4 and, besides ABC, there are three more triangles: AHB, BHC, AHC, H being the orthocenter of ABC. Indeed, since AH_a is orthogonal to BC so BC is orthogonal to AH. (In general terms, the relation of orthogonality is symmetric and instead of claiming that one line is orthogonal to another we may simply say that the two lines are orthogonal.) Again, by the symmetry of orthogonality, AC serves as another altitude in triangle AHB. Thus, in triangle AHB we have three altitudes: CH_c, BC, and AC that intersect at the point C, the orthocenter of AHB.

Similar considerations apply to the triangles BHC and AHC.

References

1. D.Wells, You are a Mathematician, John Wiley & Sons, 1995

Copyright © 1996-2008 Alexander Bogomolny