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:

## Takeaway

In any *right triangle*, the following statements are *equivalent*:

- the right triangle is isosceles
- the Kosnita point coincides with the centroid.

## Task

- (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 .