Complex bash with $A, B, E, F$ on the unit circle, so that $$|a|=|b|=|e|=|f|=1$$$$g = \frac{af(b+e) - be(a+f)}{af-be}$$Now let $M$ be the midpoint of $CD$. Since $GD = GC$, we have $GM\perp CD$ and so $GM \perp AB$. Thus $$\frac{g-m}{a-b} = -\frac{\overline{g}-\overline{m}}{\overline{a}-\overline{b}}$$$$\frac{abf+aef-abe-bef-afm+bem}{(a-b)(af-be)} = \frac{ab(a+f-b-e-af\overline{m}+be\overline{m})}{(a-b)(af-be)}$$$$abf+aef-abe-bef-afm+bem = a^2b+abf-ab^2-abe-a^2bf\overline{m}+ab^2e\overline{m}$$$$\overline{m} = \frac{m(af-be) + (a-b)(ab-ef)}{ab(af-be)}$$Since $M$ is the midpoint of $CD$, we see that $M$ lies on line $NP$, where $N$ is the midpoint of $AB$ and $P$ is the intersection of lines $AD$ and $BC$. So $$n = \frac{a+b}2$$$$p = \frac{ae(b+f) - bf(a+e)}{ae-bf}$$and so $$\frac{n-m}{n-p} \in \mathbb{R}$$$$\frac{(ae-bf)(a+b-2m)}{a^2e-abe+abf-b^2f-2aef+2bef} = \frac{(ae-bf)(a+b-2ab\overline{m})}{a^2e-abe+abf-b^2f-2a^2b+2ab^2}$$$$(a+b-2m)(a^2e-abe+abf-b^2f-2a^2b+2ab^2) = (a+b-2ab\overline{m})(a^2e-abe+abf-b^2f-2aef+2bef)$$$$(a+b-2m)(ae+bf-2ab) = (a+b-2ab\overline{m})(ae+bf-2ef)$$$$\overline{m} = \frac{m(ae+bf-2ab) + (a+b)(ab-ef)}{ab(ae+bf-2ef)}$$Setting these equal gives $$\frac{m(af-be) + (a-b)(ab-ef)}{ab(af-be)} = \frac{m(ae+bf-2ab) + (a+b)(ab-ef)}{ab(ae+bf-2ef)}$$$$m(af-be)(ae+bf-2ef) + (a-b)(ab-ef)(ae+bf-2ef) = m(af-be)(ae+bf-2ab) + (a+b)(ab-ef)(af-be)$$$$2m(af-be)(ab-ef) = (ab-ef)(a^2f-b^2e-a^2e+2aef+b^2f-2bef)$$$$m = \frac{2aef-2bef-a^2e+a^2f-b^2e+b^2f}{2(af-be)}$$Now we find the coordinate of $C$. Since $C$ lies on line $BF$, we have $$\overline{c} = \frac{b+f-c}{bf}$$Since $CM\parallel AB$, we have $$\frac{c-m}{a-b} \in \mathbb{R}$$$$\frac{2c(af-be) - (2aef-2bef-a^2e+a^2f-b^2e+b^2f)}{2(a-b)(af-be)} = -\frac{2ab\overline{c}(af-be) - (2a^2b-2ab^2-a^2e+a^2f-b^2e+b^2f)}{2(a-b)(af-be)}$$$$2c(af-be) - (2aef-2bef-a^2e+a^2f-b^2e+b^2f) = -2ab\overline{c}(af-be) + (2a^2b-2ab^2-a^2e+a^2f-b^2e+b^2f)$$$$2c(af-be) + 2ab\overline{c}(af-be) = (2a^2b-2ab^2-2a^2e+2a^2f-2b^2e+2b^2f+2aef-2bef)$$$$c(af-be) + ab\overline{c}(af-be) = a^2b-ab^2-a^2e+a^2f-b^2e+b^2f+aef-bef$$$$cf(af-be) + a(b+f-c)(af-be) = a^2bf-ab^2f-a^2ef+a^2f^2-b^2ef+b^2f^2+aef^2-bef^2$$$$c(a-f)(af-be) = a^2ef+ab^2f-ab^2e-abef-aef^2+b^2ef-b^2f^2+bef^2$$$$c(a-f)(af-be) = (a-f)(aef+b^2f-b^2e-bef)$$$$c = \frac{aef-bef-b^2e+b^2f}{af-be}$$Similarly, $$d = \frac{aef-bef-a^2e+a^2f}{af-be}$$and we see that $$|g-c| = |g-d| = \frac{|a-b||e-f|}{|af-be|}$$Now we do the last part of the problem in reverse. It will suffice to define $H$ to be the orthocenter of $\triangle ABG$, and then show that $H$ lies on the circumcircles of both $\triangle ADG$ and $\triangle BCG$. Now $H$ is the intersection of the altitude from $A$ to $BE$, and the altitude from $B$ to $AF$. So $$h = \frac{a\left(-\frac{be}a\right)\left(b-\frac{af}b\right) - b\left(-\frac{af}b\right)\left(a-\frac{be}a\right)}{a\left(-\frac{be}a\right) - b\left(-\frac{af}b\right)} = \frac{a^2f-b^2e+aef-bef}{af-be}$$Now we have the vectors \begin{align*} a-h &= -\frac{e(a-b)(b+f)}{af-be} \\ d-g &= -\frac{a(a-b)(e-f)}{af-be} \\ a-g &= \frac{f(a-b)(a-e)}{af-be} \\ d-h &= -\frac{e(a+b)(a-b)}{af-be} \end{align*}and so $$\frac{(a-h)(d-g)}{(a-g)(d-h)} = -\frac{a(b+f)(e-f)}{f(a+b)(a-e)}$$which is real. Similarly, $$\frac{(b-h)(c-g)}{(b-g)(c-h)} = -\frac{b(a+e)(f-e)}{e(a+b)(b-f)}$$which is real. $\blacksquare$

Let \( A' \) and \( B' \) be the reflections of \( A \) and \( B \) about the perpendicular bisector of \( CD \). Notice that \( B' \) lies on the circumcircle of \( \triangle ADG \) since \[ \angle GB'D = \angle GBC = \angle GBF = \angle GAE = \angle GAD. \]Similarly, \( A' \) lies on the circumcircle of \( \triangle BCG \). By symmetry, \( H \) also lies on the perpendicular bisector of \( BC \). Thus, \( HG \perp AB \). It suffices to show that \[ 180^\circ - \angle AGB = 180^\circ - \angle A'GB' = \angle GA'B' + \angle GB'A' = \angle GHB + \angle GHA = \angle AHB. \]

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