In the diagram below, $CD = BF = 10$ units and $\angle CED = \angle BAF = 30^\circ$. What would be the area of $\triangle AED$?
Show Hint
When a fold/height makes $30^\circ\!-\!60^\circ\!-\!90^\circ$ triangles, pin down the two legs quickly from the short leg $k$: the long leg is $k\sqrt{3}$ and the hypotenuse is $2k$. Use perpendiculars as altitudes to split the big area cleanly.
Step 1: Work with $\triangle CED$.
At $D$ there is a right angle and $\angle CED=30^\circ$. Hence $\,\triangle CED\,$ is a $30^\circ\!-\!60^\circ\!-\!90^\circ$ triangle with the short leg $CD=10$.
Therefore
\[
ED=10\sqrt{3},\qquad CE=2\cdot 10=20.
\]
Consequently
\[
[\triangle CED]=\tfrac12\cdot CD\cdot ED=\tfrac12\cdot 10\cdot 10\sqrt{3}=50\sqrt{3}.
\]
Step 2: Use $\angle BAF=30^\circ$ in $\triangle AEF$.
The segment $BF$ is perpendicular to $AE$ (shown in the figure), so $BF$ is the altitude from $F$ to the side $AE$ of $\triangle AEF$.
With $\angle BAF=30^\circ$ and $BF=10$ we have
\[
\tan 30^\circ=\frac{BF}{AB}=\frac{10}{AB}\ \Rightarrow\ AB=\frac{10}{\tan 30^\circ}=10\sqrt{3}.
\]
Step 3: Area of $\triangle AEC$.
Since $CE$ is perpendicular to $AE$ (from Step 1), $CE$ is the altitude to the base $AE$ in $\triangle AEC$. Thus
\[
[\triangle AEC]=\tfrac12\cdot AE\cdot CE=\tfrac12\cdot AE\cdot 20=10\,AE.
\]
Also $AE=AB+BE$, and the vertical drop from $B$ to $E$ equals $CE$ (same perpendicular), hence $BE=CE=20$.
So
\[
AE=AB+BE=10\sqrt{3}+20
\quad\Longrightarrow\quad
[\triangle AEC]=10(10\sqrt{3}+20)=100\sqrt{3}+200.
\]
Step 4: Area of $\triangle AED$.
The big triangle splits into two right-triangle parts along $CE$:
\[
[\triangle AED]=[\triangle AEC]+[\triangle CED]
=(100\sqrt{3}+200)+(50\sqrt{3})
=50(\sqrt{3}+4).
\]
\[
\boxed{[\triangle AED]=50\bigl(\sqrt{3}+4\bigr)}
\]
