If the title of the post seems obscure, look here.

I have another geometry problem for you. But before I get to the really interesting question, I'll just start with the basics. Here is a triangle with sides as labeled and altitudes as drawn.

(Almost) any schoolchild knows the formula for the area of a triangle: one-half the base times the height. A geometric proof of this is easy to construct:

The total area of the two rectangles is twice the area of the triangle, because the diagonals cut each rectangle in half. The total area is:

a * h + a' * h = A * h

We divide by 2 to get the area of the triangle.

Since the formula works

*no matter which side we choose*, there must be a connection between each side and its altitude, compared to the other pairs. Consider the altitudes for side A (AH) and side B (BH). Label the angle opposite side C as γ.

Notice there are two right triangles with γ as an angle. One has side BH, while the other has side AH. We construct two ratios which are both equal to sin(γ), since both ratios are equal to the same thing, they are equal to each other.

Easy, huh? What got my interest (and might interest you too) is that when we draw all three altitudes, they seem to cross at a single point. But can we prove this? I drew the figure below to label all the angles. In this figure, we have altitude AH perpendicular to A, and CH perpendicular to C, and then draw BH through the point where AH and CH cross. The problem is to prove that the angles where BH meets B are right angles.

I've tried for a long time to work through this, without getting anywhere. But I found the answer through another approach. As it happens, the point in the center where the altitudes cross does exist, it is called the orthocenter. Proving that the orthocenter has the properties we require lies in consideration of another special point of the triangle called the circumcenter. That is for next time.

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