Let $D$ be an interior point of the acute triangle $ABC$ with $AB>AC$ such that $\angle DAB=\angle CAD$. The point $E$ on the segment $AC$ satisfies $\angle ADE=\angle BCD$, the point $F$ on the segment $AB$ satisfies $\angle FDA=\angle DBC$, and the point $X$ on the line $AC$ satisfies $CX=BX$. Let $O_1$ and $O_2$ be the circumcenters of the triangles $ADC$ and $EXD$, respectively.

Prove that the lines $BC$, $EF$, and $O_1{O}_2$ are concurrent.

Step 1 (AD is the angle bisector): $\angle DAB=\angle CAD$ means $AD$ bisects $\angle BAC$. By the angle bisector theorem, $\frac{BD}{DC}=\frac{AB}{AC}$.

Step 2 (Identify cyclic quadrilaterals): The condition $\angle ADE=\angle BCD$ implies that $A,D,E,C$ relate through a cyclic configuration (or an angle-angle similarity). Similarly $\angle FDA=\angle DBC$ relates $F$ to the circumcircle of $ABD$.

Step 3 (Point $X$): $X$ is on line $AC$ with $CX=BX$, so $X$ lies on the perpendicular bisector of $BC$ intersected with line $AC$.

Step 4 (Circumcenters $O_1$, $O_2$): $O_1$ is the circumcenter of $△ADC$ and $O_2$ is the circumcenter of $△EXD$. The line $O_1{O}_2$ is perpendicular to... the radical axis of the two circumcircles (if they share chord $D...$).

Step 5 (Concurrence): Using the angle conditions, show that the intersection of $BC$ and $EF$ lies on the radical axis of the circumcircles of $ADC$ and $EXD$, which forces $O_1{O}_2$ to pass through this point. This is established via a power-of-a-point computation at the intersection.

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