How would I prove that BCFH is cyclic?
ABCD is parallelogram E, F Always lie within ABCD. AECF is a parallelogram aswell. H is the Intersection (other than A) of the Circumcircles of the triangles AFD and ABE.
I have tried to prove that the Diagonals of BHFC are the same, by trying to find relations between Angles (for example you can say that due the triangles ABE and DFC being congruent the angles ∠DFC and ∠AEB are the same, furthermore since H is on the same circle as E the inscribed formula states that ∠AHB = ∠AEB = ∠DFC.) but I've been out of luck.
Im currently trying the approach of defining each point by two variable coordinates and then, using the Formulas for circumcircles, trying to find an Intersection between $\odot(BFC)$, $\odot(AEB)$ and $\odot(AFD)$. But this approach is obviously very tedious and prone to errors and id still like to know if there's a geometric way to prove it.
For the configuration shown below, define $\alpha := \angle BCF$ and $\beta:=\angle ADF$ ...
$$\begin{align} \angle EAD = \alpha &\qquad \angle EBC = \beta &&(\text{parallels})\tag1 \\[4pt] \angle AEB &= \alpha+\beta &&(\text{parallels})\tag2 \\[4pt] \angle AHB &= \alpha+\beta &&(\text{Inscribed Angle Theorem in $\bigcirc AEB$}) \tag3 \\[4pt] \angle AHF &= 180^\circ-\beta &&(\text{$\square AHFD$ is cyclic}) \tag4 \end{align}$$
Therefore, $$\angle BHF = 360^\circ-\angle AHB-\angle AHF=180^\circ-\alpha \tag5$$ so that $\square BHFC$ is cyclic. $\square$
Line $(4)$ depends upon $A$, $H$, $F$, $D$ appearing in that order on their circle. Adjusting the argument for the order $A$, $F$, $H$, $D$ (with which $\angle AHF=\beta$) is left as an exercise to the reader. (There should be an argument that neatly avoids such case-work, but I haven't found it.)