To solve the problem and find the size of each angle, we proceed step by step based on the properties of isosceles triangles and parallel lines.

Key Information:

1. Triangle ABC is isosceles: $AB = AC$, so $\angle ABC = \angle ACB$.
2. AC \parallel DE.
3. $\angle BAC = 72^\circ$ and $\angle ACD = 37^\circ$.

Steps to Solve:

1. Find $\angle ABC$ and $\angle ACB$ in triangle $\triangle ABC$:

   • Since $\triangle ABC$ is isosceles, $\angle ABC = \angle ACB$.
   • The sum of angles in a triangle is $180^\circ$, so: $\angle BAC + \angle ABC + \angle ACB = 180^\circ$ Substituting $\angle BAC = 72^\circ$: $72^\circ + 2\angle ABC = 180^\circ$ $2\angle ABC = 108^\circ \implies \angle ABC = \angle ACB = 54^\circ$