In the quadrilateral $ABCD$, sides $AB$ and $CD$ are parallel, and side $AB$ is twice as long as side $CD$. Point $S$ is the intersection of the diagonals of quadrilateral $ABCD$, and triangles $ABS$ and $CDS$ are both equilateral. Point $E$ is a point on segment $BS$ such that the measure of angle $ACE$ is $30^{\circ}$.

Determine the ratio of the areas of quadrilateral $ABCD$ and triangle $EBC$.

(E. Semerádová)
From the problem statement, we know that (1) triangles $A B S$ and $C D S$ are equilateral and (2) the lengths of their sides are in the ratio $2: 1$. From the first fact, it follows that all internal angles in these triangles are $60^{\circ}$. Therefore, the size of angle $E S C$ is $180^{\circ} - 60^{\circ} = 120^{\circ}$. Since the size of angle $S C E$ is $30^{\circ}$, the size of angle $S E C$ in triangle $S E C$ is also $30^{\circ}$. Triangle $S E C$ is therefore isosceles with base $E C$, and thus segments $S E$ and $S C$ are congruent. From the second fact, it follows that segment $S B$ is twice the length of segment $S C$. Therefore, point $E$ lies at the midpoint of segment $S B$, or segment $D B$ is divided into thirds by points $S$ and $E$.

![](https://cdn.mathpix.com/cropped/2024_04_17_f2fcb56db57152892a4ag-2.jpg?height=626&width=456&top_left_y=1783&top_left_x=800)

Triangles $D B C$ and $E B C$ share vertex $C$ and the lengths of the sides opposite to it, $D B$ and $E B$, are in the ratio $3: 1$. Therefore, their areas are also in the same ratio,

$$
S_{D B C}: S_{E B C}=3: 1 .
$$

Trapezoid $A B C D$ is divided by diagonal $D B$ into two triangles $D B A$ and $D B C$, which share the height of the trapezoid and the lengths of the corresponding sides $A B$ and $C D$ are in the ratio $2: 1$. Therefore, the areas of these triangles are also in the same ratio, i.e.,

$$
S_{A B C D}: S_{D C B}=3: 1
$$

In total, we find that the desired ratio of areas is

$$
S_{A B C D}: S_{E C B}=9: 1
$$

Note. Triangles $A B S$ and $C D S$ are similar with a similarity coefficient of $2: 1$, so their areas are in the ratio $4: 1$. This ratio can be visually demonstrated by dividing triangle $A S B$ using the medians into four triangles congruent to $C D S$, see the figure. By adding segment $D F$, trapezoid $A B C D$ is divided into nine triangles with the same area as triangle $E B C$.

![](https://cdn.mathpix.com/cropped/2024_04_17_f2fcb56db57152892a4ag-3.jpg?height=628&width=462&top_left_y=1205&top_left_x=797)

Evaluation. 3 points for determining that point $E$ is at the midpoint of segment $S B$; 1 point for each of the highlighted ratios (or corresponding sub-steps in a different solution).