In the right-triangle $\Delta ABC\;(\measuredangle BCA=90^{\circ})$ $\overline{CD}\;(|CD|=v)$ is the altitude on the edge $\overline{AB}\;(|AB|=c)$. $x$ and $y$ are perpendicular to the edges $\overline{AC}$ and $\overline{BC}$ respectively. Prove $cxy=v^3$.
I was wondering if there are efficient methods that don't necessarily rely on Euclid's theorem for triangle or trigonometry.
Here is my approach:
Let $F\in\overline{AC}$ s.t.$|DF|=x$ and let $E\in\overline{BC}$ s.t. $|DG|=y$.
$$AC\perp DF\;\&\;AC\perp BC\implies DF\parallel BC$$ $$CD\perp BC\;\&\; BC\perp AC\implies CD\parallel AC$$ $CFDG$ is a rectangle.$\implies v^2=x^2+y^2$ $$\Delta ADF\sim\Delta FDC\sim\Delta EDB\sim ECD$$ Let $|AD|=p$ and $|BD|=q$ $$\frac{x}{p}=\frac{y}{v}\implies p=\frac{x}{y}v$$ $$\frac{y}{q}=\frac{x}{v}\implies q=\frac{y}{x}v$$ $$c=p+q=\frac{x}{y}v+\frac{y}{x}v=\frac{v^3}{xy}\implies cxy=v^3$$
Picture:
$$Q.E.D.$$
Thank you in advance!
Note that the right triangles BCD and CDG are similar. So are the right triangles ACD and CDF. Then,
$$\frac xv = \frac v{BC}, \>\>\>\>\> \frac yv = \frac v{AC}$$
Recognize
$$Area_{ABC} = \frac 12 vc =\frac12 BC\cdot AC $$
and multiply the two equations to arrive at
$$cxy =v^3$$