In the figure below, ACDE is a parallelogram and BDE is a triangle. $\angle CAE = 75^\circ$, $\angle BDC = 18^\circ$ and BE = BD. Find $\angle AEB$. Answer: ___

1. Find angle CDE using the properties of a 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 angle AED using the properties of a parallelogram.

Interior angles between parallel lines $AC$ and $ED$ (or consecutive angles of a parallelogram) add up to $180^\circ$.
$\angle AED = 180^\circ - \angle CAE = 180^\circ - 75^\circ = 105^\circ$

5. Calculate angle AEB.

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