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Kosnita point in a right triangle II

Let $ABC$

be a right triangle in which $\angle C=90^{\circ}$

, $F_c$

is the foot of the altitude from $C$

, $K$

is the Kosnita point, and $S$

is the symmedian point. Then the lengths of the segments $SK,KF_c, CK$

form a geometric progression:

$KF_c^2=(CK)(SK)$

Place the vertices of the right triangle at convenient points: $A(0,2u)$ , $B(2v,0)$ , and $C(0,0)$ . Then:

$F_c$ is the point $F_c\left(\frac{2u^2v}{u^2+v^2},\frac{2uv^2}{u^2+v^2}\right)$ ;
$K$ is the point $\left(\frac{4u^2v}{3(u^2+v^2)},\frac{4uv^2}{3(u^2+v^2)}\right)$ ;
$S$ is the point $\left(\frac{u^2v}{u^2+v^2},\frac{uv^2}{u^2+v^2}\right)$ .

Find the distance from the Kosnita point to the foot of the altitude from $C$

.

Using the given coordinates, we find:

$KF_c=\frac{2uv}{3\sqrt{u^2+v^2}}$

Find the distance from the Kosnita point to the symmedian point.

Using the given coordinates, we find:

$KS=\frac{uv}{3\sqrt{u^2+v^2}}$

Find the distance from vertex $C$

to the Kosnita point.

Using the given coordinates, we find:

$CK=\frac{4uv}{3\sqrt{u^2+v^2}}$

PROVE that $KF_c^2=(CK)(SK)$

.

Follows from

$KF_c=\frac{2uv}{3\sqrt{u^2+v^2}},~SK=\frac{uv}{3\sqrt{u^2+v^2}},~CK=\frac{4uv}{3\sqrt{u^2+v^2}}$

PROVE that $CK=2SK$

.

Follows from

$SK=\frac{uv}{3\sqrt{u^2+v^2}},~CK=\frac{4uv}{3\sqrt{u^2+v^2}}.$

Takeaway

In any right triangle, the following statements are equivalent:

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

Task

(Foot of the symmedian) Let be a triangle having vertices at , , . VERIFY that:
1. the foot of the symmedian from $C$ is $S_c\left(-\frac{6}{5},0\right)$
2. the foot of the symmedian from $C$ is equidistant from the feet of the altitudes from $A$ and $B$ .