HMIC

Solution:

We will show that $CE^2 = (CF)(CC')$. By a simple computation using the given angles, one may find that this is equivalent to $CF = AC - AB$, or $AF = 2AC - AB$.

We compute $AF$ by trigonometry. Assume for simplicity that $AC = \frac{1}{2}$, so $AD' = 2AD = 2AC = 1$ because $\triangle ACD$ is isosceles by angle chasing. We first compute $PD'$ by the law of cosines on triangle $APD'$, which yields
$$PD'^2 = AP^2 + AD'^2 - 2(AP)(AD') \cos 75 = AP^2 + 1 - 2AP \cos 75.$$
We may easily compute $AP$ by the law of sines in triangle $BAP$ to be $\frac{1}{\sqrt{2}}$. Thus,
$$PD'^2 = \frac{1}{2} + 1 - \sqrt{2} \cdot \frac{\sqrt{6} - \sqrt{2}}{4} = \frac{4 - \sqrt{3}}{2}.$$
By the law of sines within the circumcircle of $APD'F$, we have
$$\frac{FD'}{\sin 30} = \frac{PD'}{\sin 75}.$$
From this, we find that
$$FD' = \frac{PD' \sin 30}{\sin 75} = \frac{4 - \sqrt{3}}{\sqrt{2} + \sqrt{6}}.$$
Thus, we have by the law of cosines on triangle $D'AF$ that
$$AF^2 + AD'^2 - 2(AF)(AD') \cos 30 = AF^2 + 1 - \sqrt{3} AF = FD'^2 = \frac{4 - \sqrt{3}}{4 + 2\sqrt{3}}.$$
Finally, solving for $AF$ yields
$$AF = \frac{\sqrt{3} \pm (2\sqrt{3} - 3)}{2}$$
from which we indeed get $AF = \frac{3 - \sqrt{3}}{2}$ which equals $2AC - AB$.