I am looking for an alternative way to calculate the DE line segment. D and E are the points of intersection of the Euler line with the sides of the ABC triangle (sides: a, b and c)
method 01 - ratio, area and distance between point and line.
$rC=\frac{(ABC)}{(CDE)}$
$(CDE)=\frac{1}{2}·DE·hC$
$DE=\frac{2·(CDE)}{hC}$
$rC=\frac{(hC+hA)(hB+hC)}{hC^2}$
$DE=\frac{2·(ABC)·hC}{(hC+hA)(hB+hC)}$
where
$w=\sqrt{(abc)^2-(-a^2+b^2+c^2)(a^2-b^2+c^2)(a^2+b^2-c^2)}$
$hA=\left| \frac{1}{2·w}·(b^2-c^2)(-a^2+b^2+c^2) \right|$
$hB=\left| \frac{1}{2·w}·(c^2-a^2)(a^2-b^2+c^2) \right|$
$hC=\left| \frac{1}{2·w}·(a^2-b^2)(a^2+b^2-c^2) \right|$
method 02 - distance between two points. Formulas or expressions for calculation coordinates of points D and E are more laborious.
method 03 - ......?