Relations for an Orthocenter Based Midpoint Triangle.

by

Joshua Traxler

Here we will consider a triangle whose vertices are the midpoints between the original acute triangle's vertices and the orthocenter. To construct this triangle, first take an arbitraty acute triangle, ABC. Now construct the orthocenter, H, and the segments HA, HB, and HC. Construct the midpoints of HA, HB, and HC, and label them A', B', and C'. Connect the midpoints to form a triangle. We will refer to our newly constructed triangle as the 'ortho triangle' or A'B'C'.

The result should be something like this:

Now, we'd like to see what properties this triangle has in common with the original triangle and whether it is congruent to the median.

First, let's consider the triangles AHC and A'HC'. These appear to be similar triangles. They both have the same angle AHC. We also see tha segments A'H and HC' are the length of segments AH and HC. The the side-angle-side theorem, A'HC' is congruent to a triangle that has half the side lengths of AHC. Therefore, A'C' is half the length of AC. Similarly, distance(A'B')=1/2distance(AB) and distance(B'C')=1/2distance(BC). Therefore, we see that A'B'C' is congruent to a triangle with half the side lengths of ABC. The ratio of two similar triangles is the square of the ratio of their sides. Therefore, when the side lengths of any geometry figure are halved, the new area is 1/4 of the original's. Therefore, triangle A'B'C' is 1/4 the area of ABC, similar to the medial triangle. The medial triangle is also similar to ABC and 1/4 of its area. By transitivity, the medial triangle and ortho triangle are similar triangles with the same area. Therefore, the ratio of their sides is 1, so they must be congruent.

It is apparent from the picture that the two triangles share an orthocenter. This is because the sides of the triangle are parallel and each vertice of A'B'C' lies on an altitude of ABC, so the altitudes overlap and H=H'.

Now, let's construct the Centroid for both triangles, labeling them G and G', respectively.

It appears that HG forms a segment for which G' is the midpoint. We recall that triangle A'B'C' is a similar triangle with half the side lenght of ABC. Doing the same construction on similar geometric objects generates a new pair of similar objects. Therefore, triangle A'H'G' must be similar to AHG with half the side length. Therefore, the distance from G' must be half the distance from H as G is and be on the segment HG. They must be on the same line, because A and A' are on the same line and angle A'H'G' is the same as angle AHG.

We define the circumcenters as J and J' and incenters as I and I'. By the previous argument, J' will be the midpoint of segment HJ and I' will be the midpoint of segment HI.

For the Circumcenters:

And the Incenters:

In conclusion, we discovered that the 'ortho triangle' of an acute triangle congruent to the medial triangle. We also found that the 'orthotriangle' and original acute triangle have the same orthocenter, but the other centers are in the same direction from the orthocenter but half the distance away.