A "chord" of a square, subtending a $45^\circ$ angle at a vertex, determines a triangle whose area is bisected by the square's "other" diagonal The following Image shows the square ABCD, a point E on the side BC and the segment AF, which is a rotation of AE, at 45° (not necessarily congruent). DB is the diagonal of the square; G and H are the intersection points of the diagonal and the line segments AF, AE (respectively). Prove that the area of triangle AGH equals the area of quadrilateral FGHE. 

Bonus question: Trace a perpendicular to FE from point A, and let the point A' be the intersection of said perpendicular with segment line FE. Prove that A' describes a circumference with center A, when point E moves along the segment BC. 
Note: I know this can be solved with some analysis and analytic geometry, but I'd love to see a purely geometric solution. 
 A: Let $AB=a$ and $\measuredangle EAB=\alpha$.
Thus, $AF=\frac{a}{\cos(45^{\circ}-\alpha)}$ and $AE=\frac{a}{\cos\alpha}$.
Hence, $$S_{\Delta AFE}=\frac{AF\cdot AE\cdot \sin(\angle FAE)}{2}=\frac{1}{2}\cdot\frac{a}{\cos(45^{\circ}-\alpha)}\cdot\frac{a}{\cos\alpha}\sin45^{\circ}=\frac{a^2}{2\sqrt2\cos\alpha\cos(45^{\circ}-\alpha)}.$$
In another hand, by law of sines for $\triangle AGD$
$$\frac{AG}{\sin45^{\circ}}=\frac{a}{\sin(90^{\circ}+\alpha)}$$ and for $\triangle AHB$
$$\frac{AH}{\sin45^{\circ}}=\frac{a}{\sin(135^{\circ}-\alpha)},$$
which gives
$$S_{\Delta AGH}=\frac{AG\cdot AH\cdot \sin(\angle GAH)}{2}=\frac{1}{2}\cdot\frac{a\sin45^{\circ}}{\cos\alpha}\cdot\frac{a\sin45^{\circ}}{\sin(135^{\circ}-\alpha)}\cdot\sin45^{\circ}=$$
$$=\frac{a^2}{4\sqrt2\cos\alpha\cos(45^{\circ}-\alpha)}$$
and we are done!
Now, about the bonus. 
By law of cosines for $\Delta AFE$ easy to see that $$EF=\frac{a}{\sqrt2\cos\alpha\cos(45^{\circ}-\alpha)},$$
which gives $AA'=a$.
Done again! 
Another way.
Let $$\Delta AA''E\cong\Delta ABE$$ and  $$\Delta AA'''F\cong\Delta ADF.$$
Thus,$$A'\equiv A''\equiv A'''$$ and we are done!
A: Solution via angle chase. (Buckle up!)

Renaming a few things, I'll restate the setup as follows: 

Take $\square ABCD$ with $P$ and $Q$ on diagonal $\overline{BD}$ (with $P$ between $B$ and $Q$) such that $\angle PAQ = 45^\circ$. Let $H$ and $K$ be the centers of the (congruent) circumcircles of $\triangle APQ$ and $\triangle CPQ$, and let $\bigcirc K$ meet sides $\overline{BC}$ and $\overline{DC}$ at respective points $P^\prime$ and $Q^\prime$.



Claim: $A$, $P$, $P^\prime$ are collinear, as are $A$, $Q$, $Q^\prime$. (See (3) below.) So, this scenario does in fact match the problem stated.

In $\bigcirc H$, inscribed $\angle PAQ$ has measure $45^\circ$, so that central $\angle PHQ$ has measure $90^\circ$. Thus, $\triangle PHQ$ is a right isosceles triangle; by symmetry, so is $\triangle PKQ$. Consequently, $\square HPKQ$ is a right-angled rhombus whose diagonal aligns with $\overline{PQ}$; that is, it is a square with sides parallel to those of $\square ABCD$. 
Via parallelism, "isoscelism", and symmetry across $\overline{BD}$, we chase angles thusly
$$
\angle HAP 
\underbrace{\quad\cong\quad}_\text{isos} \angle HPA 
\underbrace{\quad\cong\quad}_\text{par} \angle BAP
\qquad(\implies\;\angle HAB\phantom{^\prime} = 2\;\angle HAP\;) \tag{1a}$$
$$\angle HAB \underbrace{\quad\cong\quad}_\text{sym} \angle KCB \underbrace{\quad\cong\quad}_\text{isos} \angle KP^\prime C 
\underbrace{\quad\cong\quad}_\text{par} \angle PKP^\prime \tag{1b}$$
(and likewise through $Q$s) to conclude
$$\angle PKP^\prime = 2\;\angle HAP \qquad\text{and}\qquad \angle QKQ^\prime = 2\;\angle HAQ \tag{2}$$
(Also, we have 
$$\angle APH + \angle HPK + \angle KPP^\prime = 180^\circ = \angle AQH + \angle HQK + \angle KQQ^\prime \tag{3}$$
which proves the Claim.)
We see, then, that $\triangle AHP$ and $\triangle PKP^\prime$ sub-divide into identical pairs of right triangles, as do $\triangle AHQ$ and $\triangle QKQ^\prime$. Therefore,
$$\begin{align}
|\triangle APQ| &= |\triangle AHP\phantom{^\prime}| + |\triangle AHQ\phantom{^\prime}| + |\triangle PHQ| \\
&= |\triangle PKP^\prime| + |\triangle QKQ^\prime| + |\triangle PKQ| \\
&= |\square PP^\prime Q^\prime Q|
\end{align} \tag{4}$$
as desired. $\square$

For the Bonus question, where $A^\prime$ is the foot of the perpendicular from $A$ to $\overline{P^\prime Q^\prime}$: Observe that the perpendicular must actually be the extension of $\overline{AH}$ (since the angles at $A$ and $P^\prime$ must be complementary). Consequently, as $\overline{AP^\prime}$ bisects $\angle HAB$, we have $\triangle BAP^\prime \cong \triangle A^\prime AP^\prime$, so that $\overline{AB}\cong\overline{AA^\prime}$. $\square$   
