$M$ is a point on $AB$. $BMC$ and $AMD$ are constructed such that they are both equilateral & on the same side of $AB$. The circumcircle of both triangle intersect at $M$ and $N$. Prove that $AD$ and $BC$ intersects at $N$.
Can anyone verify my solution:
It's enough to show $A,N,D$ collinear or $\angle ANB+\angle BND=\pi$. But $\angle ANB=\angle ANM+\angle MNB = \pi/3 + \pi/3 = 2\pi /3$ (because they are facing the same arc). Similarly it's easy to see that $\angle BND = \pi/3$. We are done.