By the given condition, we can also prove that $EA=EB$.The proof is completed since $O,F$ lie on the perpendicular bisector of $AC$; $O,E$ lie on the perpendicular bisector of $AB$.

We have that $\angle BAE = \angle ABE = \angle FCA = 45^{\circ}$. Since $\triangle ABE$ is isosceles, the perpendicular bisector through $O$ and $AB$ meets $AC$ at $E$. With the similar reasoning, we can prove that the perpendicular bisector through $O$ and $AC$ will meet $AB$ at $F$. Therefore, since $OE \parallel FC$ and $OF \parallel EF$, then $OEHF$ is a parallelogram.

anyone with complex numbers ?

By simple angle chasing, $\angle BAC=\angle ACF=\angle ABE=45^{\circ}$. So, $\triangle AFC$ and $\triangle ABE$ are isosceles. Let $M$ be the midpoint of $AC$. Since $\triangle AFC$ is isosceles, $FM \perp AC$. Since $M$ is the midpoint of chord $AC$, $OM \perp AC$. Hence,$ F, O, M$ are collinear and $FO\perp AC$. Since $HE \perp AC, FO\parallel HE$. Similarly, $OE\parallel FH$ so $OEHF$ is a parallelogram, as desired. $\blacksquare$

Another Solution Let $M_A,M_B,X$ be the mid-points of sides $BC,AC,AB$, respectively, It's easy to spot that, $\Delta ABE $ and $\Delta AFC$ are right isosceles triangles, Now just angle chase and let $\angle EBC=x \implies \angle B=45^{\circ}+x \text{ and } \angle C=90^{\circ}-x$ Claim 1:$\textcolor{blue}{\text{ Points } X,O,E \text{ are collinear}} $ proof(1): Since, $$E \equiv (1 : 0 : \tan C) \text{ ; } O \equiv (1:\sin 2B :\sin 2C) \text{ ; } X \equiv (1:1:0)$$Let \[\begin{vmatrix} 1 & 1 & 0 \\ 1 & \sin 2B & \sin 2C \\ 1 & 0 & \tan C \end{vmatrix} = \sin 2B \tan C - [\tan C -\sin 2C] =0 \implies \cos 2x \cos x = \cos x -\sin x \sin 2x \] Hence, $$(2\cos^2 x -1)\cos x =\cos x -2\sin^2x \cos x \implies \cos^2 x =1-\sin^2x \implies \boxed{\sin^2 x +\cos^2 x =1}$$Which is true, hence, $\boxed{X-O-E}$ Main Problem : We also have $M_BA=M_BC=M_BF \implies \angle FM_BC=90^{\circ}$ and $\angle BEA =90^{\circ} \implies OF||EH$ $\angle CFB=90^{\circ} \text{ and } AX=XB \text{, but in a right isosceles, median } EX \perp AB \implies OE||FH \implies OEFH \text{ is a paralellogram }$ Edit: Just realised that there was no need to bash to prove claim 1, there is a much better proof proof(2): Since, $\Delta ABE$ is right-isosceles $\implies EX \perp AB$, but since, $AX=XB \implies OX\perp AB \implies \boxed{ X-O-E }$

AF = CF So ∠FAC = 45 so ∠BOC = 90 so BFOEC is cyclic. ∠OFC = ∠OBC = 90 - ∠A = 45 and ∠FHB = 45 so FO || EH. ∠OEB = ∠OCB = ∠OBC = 45 and ∠FHB = 45 so FH || OE so EHFO is parallelogram.