Charlie Conway 7/28/2009

Orthocenter Triangle

This investigation will focus on a triangle formed by connecting the midpoints of segments HA, HB, and HC, where H is the orthocenter of triangle ABC.

As it turns out, **triangle DEF is similar to triangle ABC**. This can be shown through Side-Side-Side similarity. Since D is the midpoint of AH, and E is the midpoint of BH, then by the mid-segment theorem, the segment connecting them, DE, is half of the length of the AB, segment subtended by AH and BH. In the same way, since F is the midpoint of HC, then DF is half the length of CA. Also, EF is half the length of BC. So by SSS, triangle DEF is similar to triangle ABC with a ratio of 1/2.

More interesting observations come as a result of creating the centers of both triangles, ABC and DEF:

(1.) The area of triangle DEF is 1/4 the area of triangle ABC.** Proof.**

(2.) The orthocenter of ABC is concurrent with the orthocenter of DEF. **Proof.**

(3.) The orthocenters, the centroids, and the circumcenters of both triangles are all collinear.

(4.) The length from the centroid of ABC to H is twice the length from the centroid of DEF to H. The same is true for the circumcenters and incenters.

Observations (3.) and (4.) can all be proved using a singular **proof.**