- 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 foot of the altitude from