Tricky triangle with variables question

Question

In the right-angled triangle (ABC), the height divides CD and the hypotenuse AB in the ratio 3:2. Calculate the relationship between height and hypotenuse exactly, as simple as possible.

Answer 1

Note that $\triangle ACD\sim\triangle CBD$. So,

$$\frac{AD}{CD}=\frac{CD}{BD}$$

Therefore,

\begin{align} \frac{\frac{3}{5}AB}{CD}&=\frac{CD}{\frac{2}{3}AB}\\ \frac{CD^2}{AB^2}&=\frac{6}{25}\\ \frac{CD}{AB}&=\frac{\sqrt{6}}{5} \end{align}

Answer 2

If you mean $CD$ is a perpendicular on $AB$ Let Hypotenuse be $3k+2k=5k$

where $AD=3k$ and $BD=2k$ then

$$(AC)^2+(BC)^2=(5k)^2$$ $$(CD)^2+(3k)^2=(AC)^2$$ $$(CD)^2+(2k)^2=(BC)^2$$ Substituting gives

$$\frac{CD}{AB}=\frac{\sqrt6}{5}$$

Answer 3

we have $$AD=\frac{3}{5}c$$ and $$DB=\frac{2}{5}c$$ and from $$h_c^2=\frac{6}{25}c^2$$ follows $$\frac{h_c}{c}=\frac{\sqrt{6}}{5}$$