*right triangle*, the Kosnita point:

- is on the altitude through the -vertex
- internally divides this altitude in the ratio .

## Construction

Let be the circumcenter of triangle . The *Kosnita point* of is constructed as follows:

- find the circumcenter of triangle , then join to vertex
- find the circumcenter of triangle , then join to vertex
- find the circumcenter of triangle , then join to vertex

The lines , , concur at the *Kosnita point*.

If is a right triangle in which , then its circumcenter is the midpoint of , and so one of the three triangles above is degenerate, namely . To somewhat compensate for this, the *Kosnita point* in a right triangle behaves nicely.

## Calculations

Observe that the given triangle is right-angled at :

- the circumcenter of is , the midpoint of
- the circumcenter of is , call this
- the circumcenter of is , call this
- the equation of line is
- the equation of line is
- the two lines and meet at .

Thus, the *Kosnita point* of the given triangle is located at .

The same triangle in the previous example:

- the equation of side is
- the equation of the altitude through is
- thus, the foot of the altitude from is .

Same triangle as in the previous examples where we obtained as the Kosnita point.

The equation of the altitude through was obtained in the previous example as . The coordinates of the Kosnita point satisfy this equation, since

and so the Kosnita point lies on the altitude through .

We have and .

Explicit calculation gives and . Thus .

## Coordinates

In example 1 we were given , , and . So: and . The Kosnita point is then:

as before.

## Caption this:

## Takeaway

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

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

## Task

- (Harmonic division) Let be a
*right triangle*having vertices at , , . PROVE that:- the
*foot of the altitude from*is - the
*symmedian point*is - the
*Kosnita point*is - and divide
*harmonically*in the ratio .

- the