If we now reflect vertex over side we obtain the circumcenter.
It follows, in the case of the -triangle, that the reflection triangle is degenerate; in addition, it degenerates to a special line connecting the centroid, circumcenter and orthocenter, known as the Euler line.
Is the -triangle the only triangle whose reflection triangle degenerates to the Euler line?
To see how this holds, we’ll show that the length of the altitude through is half the distance from to the orthocenter . Indeed:
This shows that the reflection of vertex over side is the orthocenter .
Since , we have that and . Thus, the reflection of vertex over side is also the orthocenter.
To see this, note that since triangle is isosceles with as the apex, the circumcenter lies on the axis of symmetry through . We just have to show that the radius through is twice the altitude through . The altitude through is:
We’ve already seen this by explicitly determining the reflection triangle from the preceding three examples. Alternatively, the necessary and sufficient condition for a degenerate reflection triangle is . This condition holds in the case of the -triangle:
Let be the circumradius of the triangle. If the reflection of vertex over side is the orthocenter, then we must have , as per the equivalent statements here. Similarly, if the reflection of vertex over side is the orthocenter, then we must have . These two relations yield . Since is also equivalent to , we obtain . Thus, the triangle is the -triangle.
- (Aufbau) In triangle , let be the side-lengths, the circumradius, the circumcenter, the nine-point center, and the orthocenter. PROVE that the following statements are equivalent:
- the reflection of over is
- radius is parallel to side and radius is parallel to side
- the reflection of over is and the reflection of over is