Find a ratio of triangle's height segment

Question

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?

Answer 1

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 $a:b$.

Answer 2

I have a different answer

$$ \frac{CK}{KD} = \frac{a-b}{b} $$

so I want to clarify: how did you conclude that

$$ MN = \frac{b^2}{\sqrt{a^2+b^2}} + \frac{a^2}{2\sqrt{a^2+b^2}}$$

I have MN equals to $ \frac{ab}{2\sqrt{a^2+b^2}}$