In the pentagon ABCDE shown, angle BAE = 60° and angle BCD = 45°. AE is parallel to CD, E lies on CB extended and CE = DE.

1. Find the vertex angle of isosceles triangle CDE

Since $C, B, E$ are collinear, $\angle ECD = \angle BCD = 45^\circ$.

Given $CE = DE$, triangle $CDE$ is isosceles, so
$\angle CDE = \angle ECD = 45^\circ$.

Hence
$\angle CED = 180^\circ - 45^\circ - 45^\circ = 90^\circ.$

1. Find the angle at $B$ in triangle $ABE$, then convert it to $\angle ABC$

Since $AE \parallel CD$ and $BE$ lies on the same straight line as $CE$,
$\angle AEB = \angle ECD = 45^\circ$.

In $\triangle ABE$,
$\angle ABE = 180^\circ - 60^\circ - 45^\circ = 75^\circ.$

Since $BC$ and $BE$ are opposite rays on a straight line,
$\angle ABC = 180^\circ - 75^\circ = 105^\circ.$

1. Find the equal sides of the right-angled isosceles triangle

Since $\angle CED=90^\circ$ and $CE=DE$, let $CE=DE=x$.

Using Pythagoras' theorem in $\triangle CDE$,
$x^2+x^2=12^2$
$2x^2=144$
$x^2=72$

2. Calculate the area of triangle $CDE$

Area of $\triangle CDE$:
$\frac12 \times CE \times DE$
$=\frac12 \times x \times x$
$=\frac12 \times 72$
$=36$