Trapezoids

Let the sides the line parallel to the bases creates be $x$ , $4-x$ , $y$ , $6-y$ . Here $x\neq 4$ and $y\neq 6$ .

Since the perimeters are equal, we have:

$3+x+y=9+(4-x)+(6-y)$ $\iff 3+x+y=19-(x+y)$ $\iff x+y=8\iff y=8-x.$

We have a system of equations: $\begin{cases}y=8-x\\\frac{x}{4-x}=\frac{y}{6-y}\end{cases}\iff\frac{x}{4-x}=\frac{8-x}{x-2}$ $\iff x^2-2x=32-12x+x^2\iff x=3.2$ $\iff \frac{4-x}{x}=\frac{0.8}{3.2}=\boxed{1:4}.$

Let $r$ be the fraction of the non-parallel sides below the line.

$9+(6+4)r=3+(6+4)(1-r)$

$10r=-6+10-10r$

$20r=4$

$r=\frac{1}{5}$

Therefore, 1/5 is below the line and 4/5 is above it.

1/5:4/5 = 1:4.