4. A convex quadrilateral $A B C D$ is given with a point $E$ inside side $A B$ such that $|\angle A D E|=|\angle D E C|=|\angle E C B|$. The areas of triangles $A E D$ and $C E B$ are $18 \mathrm{~cm}^{2}$ and $8 \mathrm{~cm}^{2}$, respectively. Determine the area of triangle $E C D$.

The regional round of category C takes place

on Tuesday, April 8, 2014

so that it starts in the morning and the participants have 4 hours of pure time to solve the problems. Allowed aids are writing and drawing supplies and school MF tables. Calculators, laptops, and any other electronic aids are not allowed. Each problem can earn the participant 6 points; the score boundaries $\mathrm{k}$ for determining successful solvers and successful participants will be set centrally after evaluating the statistical results of the scores from all regions. These details will be communicated to the students before the competition begins.
4. Let the area of triangle $E C D$ be denoted as $S$. The angle $D E C$ is alternate to angles $A D E$ and $E C B$, hence $A D \| E C$ and $E D \| B C$ (Fig. 1). Triangles $E D A$ and $E D C$ share the side $E D$, so the ratio of their areas is equal to the ratio of the corresponding heights. If we denote $P, Q$, and $R$ as the perpendicular projections of vertices $A, B$, and $C$ onto the line $D E$ and let $v=|A P|, w=|B Q|=|C R|$, we get from the similar right triangles $A E P$ and $B E Q$ the proportion

$$
\frac{18}{S}=\frac{v}{w}=\frac{|A E|}{|E B|}
$$

Similarly, for triangles $E C D$ and $E C B$, we find that

$$
\frac{8}{S}=\frac{|E B|}{|A E|}
$$

(In Fig. 1, the corresponding projections are only indicated, but the calculation is the same as in the previous paragraph, just with the points $A \leftrightarrow B, C \leftrightarrow D$ and the areas of triangles $A E D$ and $B E C$ swapped.) Together, we have $S: 8=18: S$ or $S^{2}=144$, so the area of triangle $E C D$ is $S=12 \mathrm{~cm}^{2}$.

Another solution. As in the first solution, we find that $A D \| E C$ and $E D \| B C$. This implies the similarity of triangles $A E D$ and $E B C$. If we denote $k$ as the corresponding similarity ratio (Fig. 2), the areas of the respective triangles are

$$
18=\frac{1}{2} a b \sin \gamma, \quad S=\frac{1}{2} k a \cdot b \sin \gamma, \quad 8=\frac{1}{2} k a \cdot k b \sin \gamma
$$

Thus, it is clear that $18 \cdot 8=S^{2}$. For the area of triangle $E C D$, we get $S=12 \mathrm{~cm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_5dbc00200c43bdfc06bdg-4.jpg?height=628&width=633&top_left_y=925&top_left_x=720)

Fig. 2

For a complete solution, award 6 points, of which 2 points for justifying the relationships $A D \| E C$ and $E D \| B C$.