在下图中，ACDE 是一个平行四边形，BDE 是一个三角形。$\angle CAE = 75^\circ$、$\angle BDC = 18^\circ$ 且 $BE = BD$。求 $\angle AEB$。 答：___ $^\circ$

1. Find the interior angle $\angle CDE$ of the parallelogram.

In parallelogram $ACDE$, opposite angles are equal.
$\angle CDE = \angle CAE = 75^\circ$

2. Find $\angle BDE$.

$\angle BDE = \angle CDE - \angle BDC$
$\angle BDE = 75^\circ - 18^\circ = 57^\circ$

3. Find $\angle BED$ using the isosceles triangle $BDE$.

Since $BE = BD$, $\triangle BDE$ is an isosceles triangle.
Base angles are equal, so $\angle BED = \angle BDE = 57^\circ$

4. Find the interior angle $\angle AED$ of the parallelogram.

Interior angles between parallel lines $AC$ and $ED$ are supplementary.
$\angle AED = 180^\circ - \angle CAE = 180^\circ - 75^\circ = 105^\circ$

5. Calculate $\angle AEB$.

$\angle AEB = \angle AED - \angle BED$
$\angle AEB = 105^\circ - 57^\circ = 48^\circ$