## Task 3.

Given is a quadrilateral $A B C D$. Point $E$ is the intersection of side $\overline{A D}$ and the line parallel to $A B$ through $C$. If $|A B|=|B C|=3,|C D|=|A D|=5$ and $|A E|=3$, determine $|E C|$.
## Solution.

Points $B$ and $D$ are equally distant from points $A$ and $C$, so $BD$ is the perpendicular bisector of segment $\overline{AC}$, meaning points $A$ and $C$ are symmetric with respect to line $BC$. Let $P$ be the midpoint of segment $\overline{AC}$.

Let $F$ be the intersection of lines $CE$ and $BD$. It follows that $|AF|=|CF|$ because $BC$ is the perpendicular bisector of segment $\overline{AC}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_6c685c9f30d2bd573903g-7.jpg?height=859&width=1008&top_left_y=1561&top_left_x=521)

Since lines $CE$ and $AB$ are parallel, $\varangle BAC = \varangle ACE$. From this, we conclude that the isosceles triangles $ABC$ and $ACF$ have the same base and equal angles, so they are congruent by the $\mathrm{K}-\mathrm{S}-\mathrm{K}$ (angle-side-angle) theorem. Therefore, $|AF|=|FC|=3$.

Additionally, because lines $EF$ and $AB$ are parallel, triangles $DEF$ and $DAB$ are similar, so we have

$$
|DE|:|EF|=|DA|:|AB|
$$

Since $|DE|=|AD|-|AE|=5-3=2$, it follows that $|EF|=\frac{6}{5}$.

Thus, we have

$$
|EC|=|EF|+|FC|=\frac{6}{5}+3=\frac{21}{5}
$$