Regular Pentagon with Exterior Triangle - find angle This is a 5 point question in a math competition.  I managed to get the 7 point questions right but got stumped by this one:
In the given figure, ABCDE is a regular pentagon. $$AD=AF$$ and $$\angle DAF = 56^\circ$$  What is $x$?

I suspect that they told us $$AD = AF$$ because they want us to consider triangle ADF as isosceles, i.e. $$\angle ADF = \angle AFD = 62^\circ$$.
Here is what I have done so far - still missing some info before I can get to $x$.

Any guidance would be greatly appreciated -thanks
 A: You have made a good start, with everything you've shown being correct, but rather than adding the line $DF$ as you did, it's more useful to add $AC$ instead to the original diagram, as shown below,

Since $|AE| = |ED|$, then $\triangle AED$ is isosceles. Since $\measuredangle DEA = \frac{3(180^{\circ})}{5} = 108^{\circ}$ then, as you've shown,
$$\measuredangle ADE = \measuredangle DAE = \frac{180^{\circ} - 108^{\circ}}{2} = 36^{\circ}$$
Due to the symmetry of a regular pentagon, $|AD| = |AC|$, so $\triangle ACD$ is also isosceles. Thus, as you've also already determined,
$$\measuredangle ADC = \measuredangle ACD = 108^{\circ} - 36^{\circ} = 72^{\circ}$$
This means
$$\measuredangle CAD = 180^{\circ} - 2\times 72^{\circ} = 36^{\circ} \;\; \to \;\; \measuredangle CAF = 56^{\circ} - 36^{\circ} = 20^{\circ}$$
Next, since $|AD| = |AC| = |AF|$, then $\triangle ACF$ is isosceles. Therefore,
$$\measuredangle AFC = \measuredangle ACF = \frac{180^{\circ} - 20^{\circ}}{2} = 80^{\circ}$$
In addition,
$$\measuredangle ACB = \measuredangle EDA = 36^{\circ}$$
Thus, we finally get
$$x = \measuredangle ACF - \measuredangle ACB = 80^{\circ} - 36^{\circ} = 44^{\circ}$$
