I2.4 In the figure, $E B C$ is an equilateral triangle, and $A, D$ lie on $E B$ and $E C$ respectively. Given that $A D / / B C, A B=C D=R$ and $A C \perp$ $B D$. If the area of the trapezium $A B C D$ is $S$, find the value of $S$.
$$
\begin{array}{l}
\angle A B C=\angle B C D=60^{\circ}, A C \text { intersects } B D \text { at } J, A C \perp B D . \\
\triangle A C D \cong \triangle D B A \text { (S.A.S.) } \\
A C=B D \text { (corr. sides, } \cong \triangle \text { 's) } \\
\angle A B D=\angle D C A \text { (corr. } \angle \mathrm{s}, \cong \triangle \text { 's) } \\
\therefore \angle J B C=60^{\circ}-\angle A B D=\angle J C B
\end{array}
$$
$\triangle J B C$ is a right-angled isosceles triangle.
$$
\begin{array}{l}
\angle J B C=\angle J C B=45^{\circ} \\
B J=C J=y, A J=A C-y=B D-y=D J=x \\
\angle J A D=\angle J D A=45^{\circ}, \angle A D C=120^{\circ}
\end{array}
$$

Apply sine formula on $\triangle A C D, \frac{56}{\sin 45^{\circ}}=\frac{x+y}{\sin 120^{\circ}}$
$$
x+y=28 \sqrt{6}, \text { area }=\frac{1}{2}(x+y)^{2} \sin 90^{\circ}=\frac{1}{2}(784)(6)=2352
$$

Method $2 \angle E B C=60^{\circ}=\angle E C B$
Draw $A F \perp B C$ and $D G \perp B C$, cutting $B C$ at $F$ and $G$.
$$
\begin{array}{l}
B F=C G=56 \cos 60^{\circ}=28 \\
A F=D G=56 \sin 60^{\circ}=28 \sqrt{3}
\end{array}
$$
$A C$ intersects $B D$ at $J, A C \perp B D$.
$$
\begin{array}{l}
\triangle A C D \cong \triangle D B A \text { (S.A.S.) } \\
\angle A B D=\angle D C A \text { (corr. } \angle \mathrm{s}, \cong \triangle \text { 's) } \\
\therefore \angle J B C=60^{\circ}-\angle A B D=\angle J C B
\end{array}
$$
$\triangle J B C$ is a right-angled isosceles triangle.
$$
\begin{array}{l}
\angle J B C=\angle J C B=45^{\circ} \\
C F=A F \cot 45^{\circ}=28 \sqrt{3} \\
F G=C F-C G=28(\sqrt{3}-1)=A D
\end{array}
$$

Area of the trapezium $A B C D=S=\frac{28(\sqrt{3}-1) \times 2+28 \times 2}{2} \cdot(28 \sqrt{3})=2352$