Equilateral Triangles on Sides of a Parallelogram III

Let $D,\;$ $E,\;$ $F,\;$ be the midpoints of $AQ,\;$ $AK,\;$ and $B_1C_1,\;$ respectively.

From the Fundamental Theorem of Directly Similar Figures, $\Delta DEF\;$ is equilateral. In addition, since $AC_1MB_1\;$ is a parallelogram, $F\;$ is the midpoint of $AM.\;$ By a homothety with center $A,\;$ $\Delta QKM\;$ is also equilateral.

By angle chasing, points $A,\;$ $N,\;$ $Q,\;$ $K,\;$ and $P\;$ are concyclic. Indeed, by the construction, $\angle KAN=60^{\circ},\;$ while $\angle KQN\;$ may be either $60^{\circ}\;$ or $180^{\circ},\;$ depending on whether $Q\;$ falls inside or outside $\Delta ABC.\;$ This makes points $A,\;$ $N,\;$ $Q,\;$ and $K,\;$ concyclic. Similarly, points and $A,\;$ $P,\;$ $Q,\;$ and $K,\;$ are also concyclic so that all five lie on circle $(AKQ).\;$ It follows that $KQ\parallel NP,\;$ making $\Delta MNP\;$ equilateral.

Points $B\;$ and $C\;$ play rather superficial roles in the formulation of the problem. A more direct setting would start with a parallelogram $AC_1MB_1$ or even $\Delta AB_1C_1.$

Indeed, construction of equilateral triangles on the sides of a parallelogram has been previously looked into and - in the present setting - would show that $\Delta QKM\;$ is equilateral. Furthermore, the above problem repeats Paul Yiu's construction of equilateral triangle related to a given parallelogram, but offers an independent proof of that result.

The problem and the solution are by Thanh Oai Dao.