D/2 The area of square $A B C D$ is $196 \mathrm{~cm}^{2}$. Point $E$ is inside the square, at the same distances from points $D$ and $C$, and such that $\angle D E C=150^{\circ}$. What is the perimeter of $\triangle A B E$ equal to? Prove your answer is correct.
D/2 Solution 1: Since the area of square $A B C D$ is $196=14^{2}\left(\mathrm{~cm}^{2}\right)$, then the side of square $A B C D$ is $14 \mathrm{~cm}$.
We claim that $\triangle A B E$ is equilateral. To prove this, we make a reverse construction, starting from an equilateral $\triangle A B E^{\prime}$, building up to square $A B C D$, and eventually showing that points $E$ and $E^{\prime}$ coincide. Thus,

Step 1. Let $E^{\prime}$ be a point inside square $A B C D$ such that $\triangle A B E^{\prime}$ is equilateral. Hence, in particular, $\angle A B E^{\prime}=60^{\circ}$.

Step 2. Then $\angle E^{\prime} B C=90^{\circ}-60^{\circ}=30^{\circ}$.
Step 3. Since $B E^{\prime}=A B=B C$ ( $\triangle A B E^{\prime}$ is equilateral and $A B C D$ is a square), we conclude that $\triangle C E^{\prime} B$ is isosceles with $B E^{\prime}=B C$ and $\angle E^{\prime} B C=30^{\circ}$.

Step 4. This in turn implies that both base angles of $\triangle C E^{\prime} B$ are $\frac{1}{2}\left(180^{\circ}-30^{\circ}\right)=75^{\circ}$. In particular, $\angle B C E^{\prime}=75^{\circ}$. (Note: Since $\angle B C E^{\prime}0) .
$$

Now back to our problem. Drop a perpendicular from point $E$ to side $B C$ of the square and mark by $H$ the foot of this perpendicular. Since $E$ is at the same distances from $C$ and $D$, it easily follows that $E$ is half-way between parallel lines $A D$ and $B C$; i.e., $E H=\frac{1}{2} A B=7 \mathrm{~cm}$.
Just as above, from isosceles $\triangle D E C$ we know that $\angle E C D=15^{\circ}$. This means that $\angle C E H=15^{\circ}$ (EH\|DC,CE is a transversal, and $\angle E C D$ and $\angle C E H$ are alternating interior angles). Thus, from right $\triangle E H C$ we can calculate $C H=7 \tan 15^{\circ}$. This implies that $B H=B C-C H=14-7 \tan 15^{\circ}$.

We are ready to calculate $\cot \angle E B C$ from right $\triangle E H B$ :
$$
\cot \angle E B C=\cot \angle E B H=\frac{B H}{E H}=\frac{14-7 \tan 15^{\circ}}{7}=2-\tan 15^{\circ}=2-(2-\sqrt{3})=\sqrt{3} .
$$

But $\cot 30^{\circ}=\sqrt{3}$, so the acute $\angle E B C$ must be $30^{\circ}$. We proved what we aimed for!

To finish off the problem, note that $\angle A B E=90^{\circ}-\angle E B C=90^{\circ}-30^{\circ}=60^{\circ}$. By symmetry, $\angle A B E=60^{\circ}$ and $\triangle A B E$ is equilateral. Thus, its perimeter is $3 A B=42 \mathrm{~cm}$.

Solution 3: It is not hard to compute the ratios of a 15-75-90 right triangle without any trig, provided one knows the ratios of the 30-60-90 triangle. Just divide the 75-degree angle into 15 degree and 60-degree angles, dissecting the original triangle into a 15-15-150 and 30-60-90. Quite elementary from there.