The diagram below is made up of a rectangle AGHB, an equilateral triangle AFG, a rectangle ADEF, and a parallelogram ABCD. Find the degree measure of ∠ABC. For diagram go to http://www.purplecomet.org/welcome/practice, the 2015 middle school contest, and go to #2
1. **Identify the angles in the given shapes:**
   - The rectangle \(ABHG\) has all angles equal to \(90^\circ\).
   - The rectangle \(AFED\) also has all angles equal to \(90^\circ\).
   - The equilateral triangle \(AFG\) has all angles equal to \(60^\circ\).

2. **Calculate the sum of angles around point \(A\):**
   - Since \(ABHG\) and \(AFED\) are rectangles, the angles at \(A\) from these rectangles are \(90^\circ\) each.
   - The angle at \(A\) from the equilateral triangle \(AFG\) is \(60^\circ\).
   - Let \(\angle BAD\) be the unknown angle at \(A\) from the parallelogram \(ABCD\).

   Therefore, the sum of the angles around point \(A\) is:
   \[
   90^\circ + 90^\circ + 60^\circ + \angle BAD = 360^\circ
   \]

3. **Solve for \(\angle BAD\):**
   \[
   90^\circ + 90^\circ + 60^\circ + \angle BAD = 360^\circ
   \]
   \[
   240^\circ + \angle BAD = 360^\circ
   \]
   \[
   \angle BAD = 360^\circ - 240^\circ
   \]
   \[
   \angle BAD = 120^\circ
   \]

4. **Use the properties of the parallelogram \(ABCD\):**
   - In a parallelogram, opposite angles are equal, and adjacent angles are supplementary (i.e., they add up to \(180^\circ\)).
   - Therefore, \(\angle BAD + \angle ABC = 180^\circ\).

5. **Calculate \(\angle ABC\):**
   \[
   \angle BAD + \angle ABC = 180^\circ
   \]
   \[
   120^\circ + \angle ABC = 180^\circ
   \]
   \[
   \angle ABC = 180^\circ - 120^\circ
   \]
   \[
   \angle ABC = 60^\circ
   \]

The final answer is \(\boxed{60^\circ}\).