Suppose you are given a triangle, and construct the following within it: beginning at any point on one of the sides, you construct a line segment parallel to one of the other sides, and ending on the third side. From this new point, you repeat the procedure. Looking only at our construction method, we can predict the behavior of this contruction.
First, we'll make an assumtion, and it's one that we'll see later to not matter. Let's assume that the line we construct ends on the side that is closest to out intial point, and parallel to the side farthest from it. Then we have just constructed a similar triangle to the parent triangle. No big deal, but then we can very quickly realize that all of our further conctructions will lead to the creation of similar triangles, and we can, without drawing a line, predict where and how these triangles will occur.
Our first segment has created a triangle, and we are at a new point. We only have one choice as to where to go next- to the third side, on a segment parallel to the first side. Reaching this point, we have only one choice again, which is to go to side one, parallel to side two. Now we have four points... two on our first side, and one on each of the other sides. The two little triangles that we've constructed are not only similar to the parent, but are congruent to one another: think about how the segments intersect. Pairs of parallel segments are intersecting other pairs of parallel segments, so the resulting parallel segments are of the same length, and of course, angles are presereved at each step. If we want to get very clever, we may skip over to our third corner, and construct a congruent triangle the only possible way. On each of the two lines forming this corner, we already have a point. Construct a new point on each side, with this point being the same distance from the intersection as the other point is from its corner. Connect the dots, and our construction is complete. Six segments... three to form the new sides of our smaller triangles, and three to get from one corner to the next.
Step by Step:
There's only one special case we have to think about... what if we start at the midpoint of one of the sides. Then we have half the work!
We start with a triangle, a given point, and we connect the vertices of the triangle with the point, extending them to the other sides:
Let's explore:
We'll focus on one side at a time, and see if we can get anything meaningful. Consider CD and AD and two interior triangles, CPD and APD. The length ratio CD/AD is the same as the area ratio CPD/APD, since AC is the base for a triangle with the altitude P.
The same is true for CD/AD and triangles CBD and ABD:
Colecting facts, we have CD/AD = (CBD-CPD)/(ABD-APD) = CPB/APB:
Now, we repeat the process for the two other sides of the triangle, and we end up with BF/CF = BPA/CPA and AE/BE = APC/BPC:
Then CD/AD * BF/CF * AE/BE = CPB/APB * BPA/CPA * APC/BPC
= CPB*APB*CPA/APB*CPA*CPB = 1.
Notice that we have made absolutely no restrictions on the either the constructionof the triangle, or of the locationof the point P. We may choose to locate P exterior to the circle. In this case, our construction forms a quarilateral with two sides of the triangle as sides, and the third side as an interior height (with extensions, as necessary):
However, this does nothing to change our reaoning, and so our relation still holds.
Another issue, that seems like an afterthought, is to condsider the special cases of when P is one of the special interior points of the triangle, like the circumcenter, incenter, centroid, etc. Again, our reasoning was independent of the particular choice for point P, and the relation is unaffected: