Trapezoid areas $ABCD$ is a trapezoid with $AB||CD$, $AB = 6$ and $CD = 15$. If the area of
$△AED = 30$, what is the area of $△AEB$? ($E$ is the intersection of the 2 diagonals) I got the answer using coordinate geometry but I want to know what concepts are needed to solve this problem and what properties of a trapezoid to used.
 A: First note that triangles $\triangle DEC$ and $\triangle AEB$ are similar (not congruent because length is different), because $\angle DCA = \angle CAB$ and $\angle CDB = \angle ABD$ (both opposite interior angles between the parallel lines). As such, we know that the following must hold:
$$
\frac{\triangle AEB}{\triangle DEC} = \left( \frac{AB}{CD} \right)^2 = \left( \frac{6}{15} \right)^2 = \frac{36}{225}
$$
and thus that
$$
\triangle DEC = \frac{225}{36} \triangle AEB.
$$
Let $h$ be the height of the trapezoid (i.e. the distance between $AB$ and $CD$). Then we have
$$
\triangle ABD = \triangle AEB + \triangle AED = \triangle AEB + 30 = \frac{1}{2}\cdot AB \cdot h=3h
$$
and
$$
\triangle ACD = \triangle DEC + \triangle AED = \triangle DEC + 30 = \frac{1}{2} \cdot CD \cdot h = \frac{15}{2}h
$$
Now we have three equations and three unknowns (i.e. $\triangle AEB$, $\triangle DEC$ and $h$). Solving the equations gives
\begin{align}
\triangle AEB &= 12 \\
\triangle DEC &= 75 \\
h &= 14
\end{align}
So the answer you are looking for is $\triangle AEB = 12$.
