The medial triangle and Euler line

The triangle formed by joining the midpoints of the sides of a given triangle is called medial triangle. On the Figure 3, $\Delta A'B'C'$ is the medial triangle of $\Delta ABC$. As we know the three medians AA', BB' and CC' meet at G - the centroid of the triangle and AG:GA'=BG:GB'=CG:GC'=2:1.

  

Figure 3: The medial triangle and Euler line

Other properties of the medial triangle are included in:

Problem 3. The medial triangle A'B'C' has its sides parallel to those of $\Delta ABC$, so the two triangles are similar. The ratio of similarity is 1/2. Furthermore the orthocenter of $\Delta A'B'C'$ is at the same time the circumcenter of $\Delta ABC$ (see Figure 3).

Proof. Since A', B' and C' are the mid points of the corresponding sides it is easily follows that AB=2A'B', AC=2A'C' and B'C'=2BC, so the first part of the problem is proved. For the second lets denote with O the circumcenter of $\Delta ABC$. Since OA', OB' and OC' are perpendicular to BC, AC and AB respectively it follows that O is the orthocenter of $\Delta A'B'C'$.

Problem 4. The orthocenter, centroid and the circumcenter of any triangle are collinear. The centroid divides the distance from the orthocenter to the circumcenter in the ratio 2:1.

Proof. By Problem 8 we have that the ratio between any two corresponding line segments (not merely corresponding sides) of $\Delta ABC$ and $\Delta A'B'C'$ is 2:1. In particular AH=2OA', where H is the orthocenter of $\Delta ABC$, because O is the orthocenter of the second triangle. We also have that AG=2GA'. Finally, since AH and OA' are both perpendicular to the side BC, they are parallel. Hence $\angle HAG=\angle OA'G$ so $\Delta HAG \sim \Delta OA'G$. So

$$\angle AGH=\angle A'GO.$$

This shows that the points O, G and H are collinear and HG=2GO.

Note. The line on which these three points lie is called the Euler line of the triangle.