Let $ABCD$ be a trapezoid with bases $AB$ and $CD$ such that $\measuredangle D=2 \measuredangle B$, $AD=5$ and $CD=2$ . What is the length of $AB$?

Question

Let $ABCD$ be a trapezoid with bases $AB$ and $CD$ such that $\measuredangle D=2 \measuredangle B$, $AD=5$ and $CD=2$ . What is the length of $AB$?

I know that the answer is $7$. I want to see how I would derive this answer rather clearly for contest math. I tried $\measuredangle D=120^{\circ}$.

Answer 1

Let $E$ be the point where the parallel line to $AD$ by $C$ meets $AB$, it follows that $AECD$ is a paralelogram.

Then we have $$\measuredangle EBC+\measuredangle BCE=\measuredangle AEC=\measuredangle CDA=2\measuredangle EBC$$ Then $\measuredangle BCE=\measuredangle EBC$, so the triangle $EBC$ is isosceles with $|EB|=|CE|$, also $|CE|=|DA|=5$ and $|AE|=|CD|=2$ because $AECD$ is a paralelogram. Therefore $$|AB|=|AE|+|EB|=2+5=7.$$