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# Kosnita point in a right triangle II

Let be a right triangle in which , is the foot of the altitude from , is the Kosnita point, and is the symmedian point. Then the lengths of the segments form a geometric progression: Place the vertices of the right triangle at convenient points: , , and . Then:

• is the point ;
• is the point ;
• is the point .
Find the distance from the Kosnita point to the foot of the altitude from .

Using the given coordinates, we find: Find the distance from the Kosnita point to the symmedian point.

Using the given coordinates, we find: Find the distance from vertex to the Kosnita point.

Using the given coordinates, we find: PROVE that .

Follows from PROVE that .

Follows from ## Takeaway

In any right triangle, the following statements are equivalent:

1. the right triangle is isosceles
2. the Kosnita point coincides with the centroid.

• (Foot of the symmedian) Let be a triangle having vertices at , , . VERIFY that:
1. the foot of the symmedian from is 2. the foot of the symmedian from is equidistant from the feet of the altitudes from and .