Let be a right triangle in which , and let be the foot of the altitude from . The segment contains two special points: the symmedian point (), and the Kosnita point (). We show an example where and divide harmonically.
Given with vertices located at , , and , find the symmedian point.
Observe that the given triangle is right-angled at . Further, in a right triangle, the symmedian point is the midpoint of the right-angled vertex and the foot of the altitude from the right-angled vertex. In a previous post, we found the foot of the altitude from to be . Thus, the symmedian point in this case is .
Given with vertices located at , , and , verify that and divide harmonically, where is the Kosnita point, is the symmedian point, and is the foot of the altitude from .
We have , , , and .
Explicit calculation gives
Thus and divide harmonically, since:
In any right triangle, the following statements are equivalent:
- the right triangle is isosceles
- the Kosnita point coincides with the centroid.
- (Equivalent statements) Let be a non-right triangle. PROVE that the three statements below are equivalent:
- (Equal steps) Suppose that triangle satisfies any of the three equivalent statements above. In such a triangle, let be the foot of the symmedian from , and let be the feet of the altitudes from in that order. PROVE that .