Problem 2354

Note that $AO = OC$ implies the lines $AO, XO$ are reflections of each other about the line parallel to $AC$ through $O$, which is the perpendicular bisector of $BD$. Call this line $l$. Let $P \neq X$ be the second intersection of circle $\odot BXD$ with line $XO$, and let $Z$ be the intersection of circle $\odot BXD$ with line $AO$ furthest from $A$. Consider a reflection across $l$. This maps $B$ to $D$, $AO$ to $XO$, and circle $\odot BXD$ to itself so the transformation must map $P$, the intersection of $XO$ and circle $\odot BXD$, to the intersection of $AO$ and $\odot BXD$ furthest from $A$ i.e. $Z$. Thus we have
$$\angle OZB = \angle DPO = \angle DPX = \angle DBX = 90^\circ - \angle BAC = \angle OCB$$
which implies $BOCZ$ is cyclic. Therefore the second intersection of circle $\odot BOC$ with line $AO$ lies on circle $\odot BXD$. Similarly, $Z$ lies on circle $\odot CYE$ so the two circles have common point $Z$ on $AO$.

Solution 2:

Let $B'$ be the reflection of $B$ in $AC$ and let $AO$ intersect circle $\odot OBC$ again at $Z \neq O$. Observe that $\angle B'CA + \angle ACZ = 2\angle ACB + \angle BCZ = 2\angle ACB + 2\angle BOZ = 2\angle ACB + (180^\circ - \angle AOB) = 180^\circ$ so $Z, C, B'$ are collinear.

Claim. Triangles $ZXA$ and $ZDB'$ are similar.

Proof. We have
$$\angle XAZ = \angle BAO = 90^\circ - \angle ACB = \angle CBB' = \angle BB'C = \angle DB'Z.$$
So it suffices to prove that $\frac{B'Z}{B'D} = \frac{AZ}{AX}$. To do this, first observe $\angle B'ZA = \angle CZO = \angle CBO = \angle XCB$ and $\angle AB'Z = \angle AB'C = \angle CBA = \angle CBX$.
Hence triangles $ZAB'$ and $CXB$ are similar so $\frac{B'Z}{AZ} = \frac{BC}{CX}$.

Let $F$ be the point on $\omega$ such that $AF$ is a diameter of $\omega$, and $J$ be the intersection of $DF$ with $CO$. Consider the inversion with respect to $\omega$ and use $P'$ to denote the image of a point $P$. $X'$ lies on line $CO$ and we have $\angle BX'J = \angle BX'O = \angle OBX = \angle OBA = \angle BAO = \angle BAF = \angle BDF = \angle BDJ$ so $BX'DJ$ is cyclic.

Let $K$ be the intersection of $AF$ with $BC$. Then we have $OB = OD$ and
$$\angle KBO = 90^\circ - \angle A = \angle DBA = \angle DFA = \angle DFO = \angle ODJ$$
$$\angle BOK = 2\angle OBA = 2\angle CBD = \angle COD = \angle JOD$$
Hence triangle $BOK$ and $DOJ$ are congruent. In particular $BK = DJ$ and
$$\angle KBD = \angle KBO + \angle OBD = \angle ODJ + \angle BDO = \angle BDJ.$$
Thus $BDJK$ is an isosceles trapezoid and $BX'DJK$ is cyclic. Inverting back this gives that $BXDK'$ is cyclic. Similarly $CYEK'$ is cyclic. Since $K$ lies on $AO$, $K'$ also lies on $AO$ completing the proof.