1. **Problem Statement:** We have two parallelograms ABCD and AFGD. AF and CD intersect at point E. Given that $AD = AE$ and $\angle DGF = 36^\circ$, we need to find $\angle BCD$. 2. **Key Properties and Formulas:** - Opposite sides of a parallelogram are equal and parallel. - Adjacent angles in a parallelogram are supplementary. - $AD = AE$ implies triangle ADE is isosceles with $AD = AE$. - $\angle DGF = 36^\circ$ is given. 3. **Step-by-step Solution:** - Since ABCD is a parallelogram, $AB \parallel DC$ and $AD \parallel BC$. - Because $AD = AE$, triangle ADE is isosceles with $AD = AE$. - Point E lies on line CD, so $E$ is between $C$ and $D$ or beyond. - Since AFGD is also a parallelogram, $AF \parallel DG$ and $FG \parallel AD$. - Given $\angle DGF = 36^\circ$, and since $FG \parallel AD$, $\angle GFD = 36^\circ$ corresponds to an angle related to $AD$. - Using the properties of parallelograms and isosceles triangle ADE, we deduce that $\angle BCD = 72^\circ$. 4. **Final answer:** $$\boxed{72^\circ}$$