Tricky geometry question

Question

Suppose $ABCD$ is a rhombus, and $M \in BC$, $N \in CD$. It is known that $AMN$ is an equilateral triangle. Find the angle $ABC$.

Answer 1

Let's forget the symmetric cases for a moment. Say (without loss of generality), $BM < CN$. Assume $M^* \in CD$ is the image of $M$ under reflection in diagonal $AC$. Then $MN$ is orthogonal to $AC$ and $AM^* = AM = AN = MN.$ Since $ABCD$ is a rhombus, $BD$ is perpendicular to $AC$ so $MM^*$ is parallel to $BD$. Now consider the circle $c_A$ centered at $A$ and of radius $AM$. Then $c_A$ goes through the points $M, \, N, \, M^*$. Then $$\angle \, MM^*N = \frac{1}{2} \, \angle \, MAN = \frac{1}{2} \, 60^{\circ} = 30^{\circ}$$. However, since $MM^*$ is parallel to $BD$ $$\angle \, BDC = \angle \, MM^*N = 30^{\circ}$$ and since $ABCD$ is a rhombus, $$\angle \, ABC = \angle \, CDA = 2 \, \angle BDC = 60^{\circ}$$

This is a bit rushed, but sounds more or less correct.