The figure is made up of triangle ADG and triangle JKL. AG // JK and AD = DG.

1. Identify the properties of triangle ADG

Since $AD = DG$, triangle $ADG$ is an isosceles triangle. The base angles $\angle DAG$ and $\angle DGA$ are equal.

2. Calculate $\angle p$ using the sum of angles in a triangle

The sum of angles in a triangle is $180^\circ$. Given the vertex angle $\angle ADG = 50^\circ$, we have:
$\angle p = \angle DAG = \frac{180^\circ - 50^\circ}{2} = \frac{130^\circ}{2} = 65^\circ$

1. Find the base angle of the isosceles triangle ADG.

In $\triangle ADG$, $AD = DG$, so it is an isosceles triangle.
$\angle DAG = \frac{180^\circ - 50^\circ}{2} = 65^\circ$

2. Use parallel lines to find $\angle q$.

Since $AG \parallel JK$, the corresponding angle to $\angle DAG$ along the transversal $AD$ is the angle at $C$ (inside $\triangle CDE$).
$\angle q = \angle DAG = 65^\circ$ (corresponding angles, $AG \parallel JK$)