Let $A'$ be the reflection of $A$ wrt $O$ and $M$ be the midpoint of $BC$. Now $M\in\omega$ and $\measuredangle A'CB=\measuredangle CBA'=\measuredangle CAB/2$. From $AX\cdot AY=AE\cdot AB=AD\cdot AC$ we get $C DXY$ is cyclic so $$\measuredangle CYM=\measuredangle CYX+\measuredangle XYM=\measuredangle CDB+\measuredangle XBM=\measuredangle CBA=\measuredangle CA'M$$implying that $CA'MY$ is cyclic. Hence $A'Y\perp CY$. Also, $$\measuredangle YOA'=\measuredangle YBC, \measuredangle YA'O=\measuredangle YCB\Rightarrow \triangle YOA'\sim\triangle YBC$$and by combining with $$\measuredangle AOZ=\measuredangle CAB-\measuredangle ZOB=\measuredangle CBA', \measuredangle CYA'=90^\circ=\measuredangle A'YZ$$gives that $YOZA'$ and $YBA'C$ are similar. Since $B$ and $A'$ are symmetry wrt $OZ$ and $OXDF$ is cyclic, $$\measuredangle OBZ=\measuredangle ZA'O=\measuredangle A'CB=\measuredangle ODF=\measuredangle OXF$$and we conclude $(BZ\cap FX)\in\omega$, as desired.

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Not the nicest solution but anyway... Let $H$ be midpoint of $BC$ and Let $HD$ meet $\omega$ at $K$ and $CY$ meet $\omega$ at $T$. Claim $: H$ lies on $BEO$. Proof $:$ Obviously $\angle OEB = \angle 90 = \angle OHB$. Claim $: DXYC$ is cyclic. Proof $:$ Note that $AX.AY = AE.AB = AD.AC$. Claim $: AEYC$ is cyclic. Proof $:$ Note that $\angle ACY = \angle DCY = \angle YXB = \angle YEB$. Claim $: TB || AC$. Proof $:$ Note that $\angle TBA = \angle TBE = \angle TYE = \angle EAC = \angle BAC$. Claim $: FOXD$ is cyclic. Proof $:$ Note that $\angle OFD = \angle 90 = \angle OXB$. Claim $: F,X,K$ are collinear. Proof $:$ Note that $\angle FXO = \angle FDO = \angle EDO = \angle EAO = \angle DAO = \angle DAH = \angle DHO = \angle KHO$ Now we need to prove $B,Z,K$ are collinear. I represent the following lemma and name it "Unicorn Lemma" Unicorn Lemma $:$ Let $ABC$ be such that $AB = BC$. Let $Y$ be reflection on $C$ wrt $AB$. Let $P$ be where tangents of $B$ and $C$ to $ABC$ meet and Let $X$ be projection of $B$ into line through $C$ parallel to $AY$. we have that $Y,P,X$ are collinear. Proof $:$ Let $\angle BAC = t$. Note that $\frac{\sin{BYP}}{\sin{CYP}} = \frac{BP}{CP} . \frac{\sin{PBY}}{\sin{PCY}} = \frac{\sin{3t}}{\sin{90-3t}}$ and $\frac{\sin{BYX}}{\sin{CYX}} = \frac{BX}{CX} . \frac{\sin{XBY}}{\sin{XCY}} = \frac{\sin{3t}}{\sin{90-3t}} . \frac{\sin{90+t}}{\sin{90+t}} = \frac{\sin{3t}}{\sin{90-3t}}$ so $\frac{\sin{BYP}}{\sin{CYP}} = \frac{\sin{BYX}}{\sin{CYX}}$ so $\angle BYP = \angle BYX$ and $\angle CYP = \angle CYX$ so $Y,P,X$ are collinear. Now Applying $Unicorn Lemma$ on $AOB$, if $Z'$ is where tangents of $B$ and $O$ to $ABO$ meet, then $C,Z',T$ are collinear. Now since $OZ' || AB || OZ$ and Both $Z$ and $Z'$ are on $CT$ then $Z$ is $Z'$. Now we have $\angle OBZ = \angle OAB = \angle KHO = \angle KBO$ so $K,Z,B$ are collinear as wanted. we're Done.

Let $M$ be the midpoint of $BC$ which obviously lies on $\omega$. The proof consists of 2 claims: 1st claim: $BZ$,$DM$ meet at $\omega$. Proof of 1st claim: Let $L$ be their intersection point and $P$ be the antidiametrical point of $A$ in the circumcircle of $ABC$. We have that $OZ$ is parallel to $AB$ and hence perpendicular to $BP$, so we have that $ZB=ZP$ . By power of point we have that $AX\cdot AY=AE\cdot AB=AD\cdot AC$, hence $CYXD$ is cyclic. So using the cyclic quadrilaterals we have that: $ \angle YCD =\angle YXB = \angle YMC$, hence $CD$ is tangent to $(MCY)$. Also we have that $\angle MPC =\angle ABC = \angle ACB = \angle BCD$, so $CD$ is also tangent to $(MCP)$. From the above we conclude that $P$ and $Y$ lie on the circle which passes through $C$ and $M$ and is tangent to $CD$. Therefore $MYCP$ is cyclic and hence $\angle PYC = 90$ . Now let $N$ be the midpoint of $BP$ which trivially lies at $OZ$ and at $ \omega $. We have that $ZNPY$ is cyclic as $ \angle PYZ = 90 = \angle ZNP$ and so $ \angle NZP = \angle NYP (1) $ Now using the cyclic quadrilaterals and that $ \angle PYC = 90$ we get that: $ \angle NYP = 360 - \angle PYC - \angle NYX - \angle XYC= 270 - (180- \angle XBN) - \angle XDA = 90+ \angle XBN - \angle BDA = 90 + \angle DBP - \angle BDA = 90 + ( 90- \angle ABD) - \angle BDA = 180 - \angle ABD - \angle BDA = \angle BAD = \angle BAC = \angle BOP (2)$ From $(1)$ and $(2)$ we conclude that: $ \angle BZP = 2 \angle NZP = 2 \angle NYP = 2 \angle BOP$ and hence (using $ZB=ZP$) we have that $Z$ is the circumcenter of $OBP$ and hence $ZB=ZO$. So we have (using that $ML$ is parallel to $AB$): $ \angle BLM = \angle ABL = \angle ABZ = \angle ABO +\angle OBZ= \angle ABO + \angle BOZ = \angle A/2 + \angle A/2 = \angle A = \angle BOM$ and hence $L$ lies on $\omega$, as needed. 2nd claim: $FX$, $DM$ meet at $\omega$. Proof of 2nd claim: Let $K$ be their intersection point. It is obvious that $\omega$ has diameter $BO$. Hence we have that $ \angle BXO = 90 = \angle OFD$ and so $OFDX$ is cyclic. Hence we have that : $\angle BXK = \angle FXD = \angle FOD = \angle ABC = \angle DMC = \angle KMC$ and hence $K$ lies on $\omega$, as needed. From the above claims we conclude that $ K \equiv L$ and hence $FX$ , $BZ$ indeed intersect at $\omega$.

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