Find a ratio of triangle's height segment Given a right angle triangle ABC (C = 90) and a median AM. CD is the height of the triangle and it intersects median in point K. What is the CK / KD ratio?

Let $AC=b$, $BC=a$ and $N$ be placed on $DB$ such that $MN\perp AB.$
Thus, $$MN=\frac{1}{2}CD,$$
$$AD=\frac{b^2}{\sqrt{a^2+b^2}}$$ and
$$BD=\frac{a^2}{\sqrt{a^2+b^2}}.$$
Id est, $$\frac{KD}{MN}=\frac{AD}{AN}=\frac{\frac{b^2}{\sqrt{a^2+b^2}}}{\frac{b^2}{\sqrt{a^2+b^2}}+\frac{a^2}{2\sqrt{a^2+b^2}}}=\frac{2b^2}{a^2+2b^2}.$$
Thus, $$\frac{KD}{CD}=\frac{b^2}{a^2+2b^2}$$ or
$$\frac{CD}{KD}=\frac{a^2+2b^2}{b^2}$$ or
$$1+\frac{CK}{KD}=\frac{a^2+b^2}{b^2}+1,$$ which gives
$$CK:KD=(a^2+b^2):b^2.$$
We see that the needed ratio depends on the ratio $a:b$.
