In the diagram, pentagon $A B C D E$ is symmetrical about altitude $C F$. Also, $A E=200$, $C F=80 \sqrt{3}, \angle A B C=150^{\circ}$, and $\angle B C D=120^{\circ}$. Determine the vertical distance between $A E$ and $B D$.

![](https://cdn.mathpix.com/cropped/2024_04_17_45c89f07dc9351e0476cg-2.jpg?height=344&width=504&top_left_y=1319&top_left_x=867)
Suppose that $C F$ and $B D$ cross at $X$.

Since the figure is symmetric about $C F$, then $B D$ is perpendicular to $C F$ and so is parallel to $A E$.

Let the distance from $B D$ to $A E$ be $h$; that is, $X F=h$.

Since $C F=80 \sqrt{3}$, then $C X=C F-X F=80 \sqrt{3}-h$.

Since the figure is symmetric about $C F$, then $C B=C D$ which means that $\triangle B C D$ is isosceles. Since $\triangle B C D$ is isosceles, then $\angle C B D=\frac{1}{2}\left(180^{\circ}-\angle B C D\right)=30^{\circ}$.

Since $\triangle B C X$ is right-angled at $X$, then it is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.

Using the ratios of sides in such a triangle, this means that

$$
B X=\sqrt{3} C X=\sqrt{3}(80 \sqrt{3}-h)=240-\sqrt{3} h
$$

Since $\angle A B C=150^{\circ}$ and $\angle C B D=30^{\circ}$, then $\angle A B D=\angle A B C-\angle C B D=120^{\circ}$.

Since $B D$ and $A E$ are parallel, then $\angle B A E=180^{\circ}-\angle A B D=60^{\circ}$.

We drop a perpendicular from $B$ to $T$ on $A E$.

This creates a rectangle $B X F T$ (it has three right angles and so must be a rectangle) and a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, $\triangle B A T$.

![](https://cdn.mathpix.com/cropped/2024_04_30_a838ccd738aa6ca69034g-09.jpg?height=347&width=507&top_left_y=1819&top_left_x=863)

Since $A E=200$, then $A F=\frac{1}{2} A E=100$.

Since $B X=240-\sqrt{3} h$, then $T F=B X=240-\sqrt{3} h$.

Thus, $A T=A F-T F=100-(240-\sqrt{3} h)=\sqrt{3} h-140$.

Also, $B T=X F=h$.

Finally, in $\triangle B A T$, we have $\frac{B T}{A T}=\sqrt{3}$, which gives $\frac{h}{\sqrt{3} h-140}=\sqrt{3}$ and so $h=3 h-140 \sqrt{3}$. Re-arranging, we obtain $2 h=140 \sqrt{3}$ and finally that $h=70 \sqrt{3}$.