Competition problem — Geometry — Maths Olympiad Prep

Geometry Difficulty 3.9 AMC 10/12 Prove it Japan

A point $E$ is located on the side $DA$ of a quadrilateral $ABCD$ in such a way that the lines $AB$ and $EC$ are parallel. If $AB = 3$ , $BC = 3$ , $CD = 5$ , $DE = 3$ and $EA = 2$ , determine $EC$ .
Here we denote for a line segment $XY$ its length also be $XY$ .

Solution

$\boxed{\frac{24}{5}}$
Let $F$ be the point of intersection of $BD$ and $EC$ . From $AB // EF$ , we get $AB : EF = DA : DE$ . Therefore, we have $EF = \frac{AB \cdot DE}{DA} = \frac{9}{5}$ .

We also have $AB = CB = 3$ , $AD = CD = 5$ , which imply that the triangles $ABD$ and $CBD$ are congruent as the side $BD$ is common to both. Consequently, we have $\angle ABD = \angle CBD$ , and this, together with the fact $AB // EC$ yields $\angle CFB = \angle ABF = \angle CBF$ . Therefore, we have $FC = BC = 3$ . We finally obtain $EC = EF + FC = \frac{24}{5}$ for the desired answer.

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