Let $ABCDE$ be a regular pentagon, and let $F$ be a point on $\overline{AB}$ with $\angle CDF=55^\circ$.  Suppose $\overline{FC}$ and $\overline{BE}$ meet at $G$, and select $H$ on the extension of $\overline{CE}$ past $E$ such that $\angle DHE=\angle FDG$. Find the measure of $\angle GHD$, in degrees.

[i]Proposed by David Stoner[/i]
1. **Lemma 1:** Let \( I = BD \cap CE \); then \( DF \parallel GI \). (In fact, this holds for all \( F \in AB \).)

2. **Proof of Lemma 1:** 
   - Let \( J = BE \cap DF \) and \( K = BF \cap CD \). 
   - Since \( BE \parallel CD \), it follows that 
     \[
     \frac{GB}{GJ} = \frac{CK}{CD}.
     \]
   - But 
     \[
     \frac{CK}{CD} = \frac{IB}{ID} \implies \frac{GB}{GJ} = \frac{IB}{ID}.
     \]
   - Consequently, \( DF \parallel GI \) by the Side-Angle-Side (SAS) similarity criterion. 
   - \(\blacksquare\)

3. By Lemma 1, \( \angle DGI = \angle FDG = \angle DHE \), whence quadrilateral \( DHGI \) is cyclic.

4. Since \( DHGI \) is cyclic, the opposite angles sum to \( 180^\circ \). Therefore, 
   \[
   \angle DHE + \angle DGI = 180^\circ.
   \]

5. Given \( \angle FDG = 55^\circ \), and knowing that \( \angle DGI = \angle FDG \), we have 
   \[
   \angle DGI = 55^\circ.
   \]

6. Since \( \angle DHE = \angle FDG = 55^\circ \), we can now find \( \angle GHD \):
   \[
   \angle GHD = \angle DHE + \angle FHE.
   \]

7. To find \( \angle FHE \), we note that \( \angle FHE = \angle BDG \) because \( DHGI \) is cyclic and \( \angle FHE \) is an external angle to \( \triangle DHE \).

8. Since \( \angle BDG = 36^\circ \) (as \( \angle BCD = 36^\circ \) in a regular pentagon), we have 
   \[
   \angle FHE = 36^\circ.
   \]

9. Therefore, 
   \[
   \angle GHD = \angle DHE + \angle FHE = 55^\circ + 36^\circ = 91^\circ.
   \]

10. However, we need to subtract the internal angle \( \angle FDB \) from \( 55^\circ \) to get the correct measure:
    \[
    \angle FDB = 55^\circ - 36^\circ = 19^\circ.
    \]

11. Thus, 
    \[
    \angle GHD = 19^\circ.
    \]

The final answer is \(\boxed{19^\circ}\).