# Proving That Disturbing Triangle Theorem That Isn’t Morley’s Somehow

I couldn’t leave people just hanging on that triangle theorem from the other day. Tthis was a compass-and-straightedge method to split a triangle into two shapes of equal area. The trick was you could split it along any point on one of the three legs of the triangle.

The theorem unsettled me, yes. But proving that it does work is not so bad and I thought to do that today.

The process: start with a triangle ABC. Pick a point P on one of the legs. We’ll say it’s on leg AB. Draw the line segment from the other vertex, C, to point P.

Now from the median point S on leg AB, draw the line parallel to PC and that intersects either leg AC or leg BC. Label that point R. The line segment RP cuts the triangle ABC into one triangle and another shape, normally a quadrilateral. Both shapes have the same area, half that of the original triangle.

To prove it nicely will involve one extra line, and the identification of one point. Construct the line SC. Lines SC and PC intersect at some point; call that Q. I’ve actually made a diagram of this, just below. I’ve put the intersection point R on the leg AC. All that would change if the point R were on BC instead would be some of the labels.

Here’s how the proof will go. I want to show triangle APR has half the area of triangle ABC. The area of triangle ARP has to be equal to the area of triangle ASC, plus the area of triangle SPQ, minus the area of triangle QCR. So the first step is proving that triangle ASC has half the area of triangle ABC. The second step is showing triangle SPQ has the same area as does triangle QCR. When that’s done, we know triangle APR has the same area as triangle ASC, which is half that of triangle ABC.

First. That ASC has half the area of triangle ABC. The area of a triangle is one-half times the length of a base times its height. The base is any of the three legs which connect two points. The height is the perpendicular distance from the third point to the line that first leg is on. Here, take the base of triangle ABC to be the line segment AC. Also take the base of triangle ASC to be the line segment AC. They have the same base. Point S is the median of the line segment AB. So point S is half as far from the base AC as the point B is. Triangle ASC has half the height of triangle ABC. Same base, half the height. So triangle ASC has half the area of triangle ABC.

Second. That triangle SPQ has the same area as triangle QCR. This is going to be most easily done by looking at two other triangles, SPC and PCR. They’re relevant to triangles SPQ and QCR. Triangle SPC has the same area as triangle PCR. Take as the base for both of them the leg PC. Point S and point R are both on the line SR. SR was created parallel to the line PC. So the perpendicular distance from point S to line PC has to be the same as the perpendicular distance from point R to the line PC. Triangle SPC has the same base and same height as does triangle PCR. So they have the same area.

Now. Triangle SPC is made up of two smaller triangles: triangle SPQ and triangle PCQ. Its area is split, somehow, between those two. Triangle PCR is also made of two smaller triangles: triangle PCQ and triangle QCR. Its area is split between those two.

The area of triangle SPQ plus the area of triangle PCQ is the same as the area of triangle SPC. This is equal to the area of triangle PCR. The area of triangle PCR is the area of triangle PCQ plus the area of triangle QCR.

And that all adds up only if the area of triangle SPQ is the same as the area of triangle QCR.

So. We had that area of triangle APR is equal to the area of triangle ASC plus the area of triangle SPQ minus the area of triangle QCR. That’s the area of triangle ASC plus zero. And that’s half the area of triangle ABC. Whatever shape is left has to have the remaining area, half the area of triangle ABC.

It’s still such a neat result.

Morley’s theorem, by the way, says this: take any triangle. Trisect each of its three interior angles. That is, for each vertex, draw the two lines that cut the interior angle into three equal spans. This creates six lines. Take the three points where these lines for adjacent angles intersect. (That is, draw the obvious intersection points.) This creates a new triangle. It’s equilateral. What business could an equilateral triangle possibly have in all this? Exactly.