I've tried several times to find an algebraic expression for the angles in the first figure of that post, to solve the problem directly, but didn't get anywhere. A few days ago, I came across a theorem attributed to Ceva, which makes the problem easy to solve (though not in the way I posed it). The proof is easy and the theorem is also easy to remember.
Ceva's theorem goes like this:
Pick any point P within the triangle ABC and draw lines through the vertices and P as shown in the figure. P can be any point, so the distances BX and CX are not equal but have some ratio
BX/CX = x. Using the standard formula for area of a triangle, and recognizing that the heights are equal in each case, we can show that the areas (designated as
|ABC|) of the triangles below are in the same ratio:
The second identity is illustrated in this diagram:
Therefore the difference is also in the same ratio:
As shown in this diagram:
By the same reasoning, if
y = CY / AYand
z = AZ / BZ, then:
Noticing that we have the same areas in both numerator and denominator we write:
That is, if the internal lines are concurrent (all intersect at P), then:
It's easy to remember the order of terms, just walk around the circle and fill in the ratios one by one.
The theorem also works in reverse, if this ratio is one, then the internal lines are concurrent. (The proof of this is very simple but a little subtle. See here).
Moving to our problem with the altitudes:
we use a little trigonometry to calculate the heights and bases. Label the angles as α, β and γ, then:
The ratio is constructed:
Since all the terms cancel, the ratio is one and therefore the altitudes are concurrent!