In the figure below, ABCD is a parallelogram. ECG is a straight line and BCE is an isosceles triangle. BC = CE and $\angle BCG = 100^\circ$. Find $\angle AEC$. Ans : ___ $^\circ$

1. Find $\angle BCE$ using angles on a straight line

Since $ECG$ is a straight line, $\angle BCE + \angle BCG = 180^\circ$.
$\angle BCE = 180^\circ - 100^\circ = 80^\circ$

2. Find $\angle CEB$ using the properties of the isosceles triangle $BCE$

In $\triangle BCE$, $BC = CE$, so $\angle CEB = \angle CBE$.
$\angle CEB = \frac{180^\circ - 80^\circ}{2} = 50^\circ$

3. Find $\angle AEC$ using angles on a straight line

Since $A, E, B$ lie on a straight line (side of the parallelogram), $\angle AEC + \angle CEB = 180^\circ$.
$\angle AEC = 180^\circ - 50^\circ = 130^\circ$