Reference: HKCEE MC 1982 Q51
$$
\begin{array}{l}
\angle A B D=24^{\circ} \text { (given) } \\
\angle C A B=24^{\circ} \text { (diagonals of rectangle) } \\
\angle A E B=132^{\circ}(\angle \mathrm{s} \text { sum of } \triangle) \\
\left.\angle C E D=132^{\circ} \text { (vert. opp. } \angle \mathrm{s}\right) \\
\angle C E F=60^{\circ}(\angle \text { of an equilateral triangle }) \\
\angle D E F=132^{\circ}-60^{\circ}=72^{\circ} \\
E D=E C=E F \text { (diagonals of rectangle, sides of equilateral } \triangle) \\
\therefore \triangle D E F \text { is isosceles (2 sides equal) } \\
\angle E F D=\angle E D F \text { (base } \angle \mathrm{s} \text { isos. } \triangle \text { ) } \\
d=(180-72) \div 2=54(\angle \text { s sum of isos. } \triangle)
\end{array}
$$
I4.4 $A B C D$ is a rectangle and $C E F$ is an equilateral triangle, $\angle A B D=6 c^{\circ}$, find the value of $d$.