Thursday, April 15, 2010

Ceva's Theorem and the altitude problem

In this post last summer, I was working on the problem of the orthocenter. It wasn't obvious to me that all three altitudes of a triangle actually cross at a single point (that they are concurrent). This turns out to be true, and that point is called the orthocenter. In the next post, I solved the problem by constructing a small triangle inscribed into a larger one, using the midpoints of the sides of the large triangle as the vertices of the small one. Remarkably, the orthocenter of the small triangle is the circumcenter of the larger one. Because we know that any three points lie on a unique circle, which has a unique center, we deduce that the circumcenter of the large triangle (and orthocenter of the small triangle) exists and it is unique.

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:

|ABX| / |ACX| = x
|BPX| / |CPX| = x

The second identity is illustrated in this diagram:

Therefore the difference is also in the same ratio:

|ABP| / |ACP| = x

As shown in this diagram:

By the same reasoning, if y = CY / AY and z = AZ / BZ, then:

|BCP| / |ABP| = y
|ACP| / |BCP| = z

Noticing that we have the same areas in both numerator and denominator we write:

       |ABP| |BCP| |ACP|
xyz = ----- ----- ----- = 1

That is, if the internal lines are concurrent (all intersect at P), then:

-- -- -- = 1

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:

AZ = AC cos α

The ratio is constructed:

AC cos α AB cos β BC cos γ
-------- --------- ---------
BC cos β AC cos γ AB cos α

Since all the terms cancel, the ratio is one and therefore the altitudes are concurrent!

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