so hmm notice the picture below
dropping a line from the right-angle, perpendicular to the hypotenuse line, produces, 3 similar triangles, a Large one, containing the two small ones, a Medium and a Small one
now, using proportions, we can use those sides as [tex]\bf \cfrac{large}{small}\qquad \cfrac{3+x}{6}=\cfrac{6}{3}\impliedby \textit{solve for "x"}
\\\\\\
\cfrac{medium}{large}\qquad \cfrac{x}{y}=\cfrac{y}{3+x}\impliedby \textit{solve for "y"}
\\\\\\
\cfrac{medium}{small}\qquad \cfrac{z}{3}=\cfrac{x}{z}\impliedby \textit{solve for "z"}[/tex]
once you get "x", you just plug that in the 2nd proportion to get "y"
and you can also use the value of "x" to get "z" on the last proportion
