In $\triangle$ABC shown below, AB = AC, AF = EF, and EH = CH = DH = GH = DG = BG. Also,
∠CHE = ∠F GH. Find the degree measure of ∠BAC.
[center][img]https://i.snag.gy/ZyxQVX.jpg[/img][/center]
1. Let us call $\angle ABC = x^{\circ}$. Since $GH = HD = DG$, we see that $\triangle GHD$ is equilateral, which means that $\angle HDG = 60^{\circ}$.
2. In $\triangle BDG$, we have:
   \[
   \angle BDG + \angle HDG + \angle HDC = 180^{\circ}
   \]
   Substituting the known values:
   \[
   x^{\circ} + 60^{\circ} + \angle HDC = 180^{\circ}
   \]
   Solving for $\angle HDC$:
   \[
   \angle HDC = 180^{\circ} - x^{\circ} - 60^{\circ} = (120 - x)^{\circ}
   \]
3. Since $\angle HDC = \angle HCD$ (because $\triangle HDC$ is isosceles), we can find $\angle ECH$:
   \[
   \angle ECH = \angle HCD = (120 - x)^{\circ}
   \]
4. Now, we calculate $\angle EHC$:
   \[
   \angle EHC = 180^{\circ} - 2 \times \angle ECH = 180^{\circ} - 2 \times (120 - x)^{\circ} = 180^{\circ} - (240 - 2x)^{\circ} = (420 - 4x)^{\circ}
   \]
5. Next, we calculate $\angle FGH$:
   \[
   \angle FGH = 180^{\circ} - \angle HGD - \angle BGD
   \]
   Since $\angle HGD = 60^{\circ}$ and $\angle BGD = 180^{\circ} - 2x$:
   \[
   \angle FGH = 180^{\circ} - 60^{\circ} - (180^{\circ} - 2x) = 180^{\circ} - 60^{\circ} - 180^{\circ} + 2x = (2x - 60)^{\circ}
   \]
6. Given that $\angle FGH = \angle EHC$, we set up the equation:
   \[
   420^{\circ} - 4x = 2x - 60^{\circ}
   \]
   Solving for $x$:
   \[
   420^{\circ} + 60^{\circ} = 6x
   \]
   \[
   480^{\circ} = 6x
   \]
   \[
   x = 80^{\circ}
   \]
7. Finally, we find $\angle BAC$:
   \[
   \angle BAC = 180^{\circ} - 2x = 180^{\circ} - 2 \times 80^{\circ} = 180^{\circ} - 160^{\circ} = 20^{\circ}
   \]

The final answer is $\boxed{20^{\circ}}$