Let $ABC$ be a triangle with circumcircle $c$ and circumcenter $O$, and let $D$ be a point on the side $BC$ different from the vertices and the midpoint of $BC$. Let $K$ be the point where the circumcircle $c_1$ of the triangle $BOD$ intersects $c$ for the second time and let $Z$ be the point where $c_1$ meets the line $AB$. Let $M$ be the point where the circumcircle $c_2$ of the triangle $COD$ intersects $c$ for the second time and let $E$ be the point where $c_2$ meets the line $AC$. Finally let $N$ be the point where the circumcircle $c_3$ of the triangle $AEZ$ meets $c$ again. Prove that the triangles $ABC$ and $NKM$ are congruent.
First note that $O$ is on $c_3$ by Miquel. Now,
$\angle BDK=\angle BOK=90-\angle OKB=90-\angle ODB=\angle ODC-90$.
$\angle MDC=\angle COM=90-\angle OMC=90-180+\angle ODC=\angle ODC -90$.
Hence $KMD$ are collinear and similarly so are $NEM$ and $NZK$. Now see that
$\angle NKM=\angle ZKD=\angle B$
And similarly $\angle KMN=\angle C$ giving $\triangle ABC \sim \triangle NKM$. Now from Reim on $c_3$ and $c$ we get that $EZ||CK$. But
$\angle NEZ=\angle NAB=\angle NMB$
gives $EZ||BM$ so we get that $CK||BM$ which means that $CKBM$ is an isosceles trapezoid. So now we have $KM=BC$ which finishes the problem.