Let $ABCD$ be a convex quadrilateral with $AD = BC$, $\angle BAC+\angle DCA = 180^{\circ}$, and $\angle BAC \neq 90^{\circ}.$ Let $O_1$ and $O_2$ be the circumcenters of triangles $ABC$ and $CAD$, respectively. Prove that one intersection point of the circumcircles of triangles $O_1BC$ and $O_2AD$ lies on $AC$.
Let $DC\cap AB=E$ and $(BCO_1)\cap AC=T$.
$\angle EAC=180-\angle DCA=\angle ECA \implies EA=EC\implies E\in O_1O_2$. Since $\frac{AD}{\sin\angle DCA}=\frac{BC}{\sin\angle BAC}\implies O_2C=O_2D=O_2A=O_1C=O_1B \implies \triangle O_2DA\sim \triangle O_1BC
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\\ \implies \angle O_2TA=\angle O_1TA=\angle O_1TC=\angle O_1BC=\angle O_2DA \implies TAO_2D$ is cyclic.So we are done.
Also we can show that second intersection point of these circles is $E$, which lies on line $AB$. Since $\angle O_1EB=90-\angle BAC=\angle O_1CB \implies E\in (BCO_1)$. Also $\angle O_2TA=\angle O_2DA=\angle O_1BC=\angle O_1CB=\angle O_1EB=\angle O_2EA \implies E\in (TAO_2D)$, as desired.