that's an equilic quadrilateral! I've got another one: let $ABCD$ a equilic quadrilateral with $\measuredangle{\text{ }}A+\measuredangle{\text{ }}B = 120^\circ$, and $\overline{AD}= \overline{BC}$. Let $O$, $P$ and $Q$ be midpoints of $\overline{AC}$, $\overline{BD}$ and $\overline{CD}$ respectively. Prove that $\triangle OPQ$ is equilateral.

The condition $AD = BC,$ is not necessary. We denote as $E,$ the intersection point of the sidelines $AD,$ $BC,$ of $ABCD.$ So, we have that $\angle CED = 60^{o}.$ From cyclic quadrilaterals $ACEP,$ $CDEQ,$ $BDER,$ we have that $\angle QEC+\angle CED+\angle AEP = 60^{o}+60^{o}+60^{o}= 180^{o}$ $\Longrightarrow$ $Q,$ $E,$ $P,$ are collinear $(1).$ $\angle QEC = \angle QDC = 60^{o}= \angle RDB = \angle REB$ $\Longrightarrow$ $R,$ $Q,$ $E,$ are collinear $(2).$ From $(1),$ $(2)$ we conclude that the points $P,$ $Q,$ $R$ are collinear and the proof is completed. Kostas Vittas.

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The contition $AD = BC,$ is necessary for the Line's problem. So, if $K,$ $L,$ $M,$ are the midpoints of the segments $AC,$ $CD,$ $BD$ respectively, we have that $\angle KLM = \angle AEB = 60^{o}$ $($ because of $KL\parallel AE$ and $LM\parallel BE$ $)$ and also $KL = \frac{AD}{2}= \frac{BC}{2}= LM.$ Hence the triangle $\bigtriangleup KLM$ is equilateral and the proof is completed. Kostas Vittas.

The triangle $BAQ$ is echilateral as well. Any nice proofs?

Line wrote: that's an equilic quadrilateral! I've got another one: let $ABCD$ a equilic quadrilateral with $\measuredangle{\text{ }}A+\measuredangle{\text{ }}B = 120^\circ$, and $\overline{AD}= \overline{BC}$. Let $O$, $P$ and $Q$ be midpoints of $\overline{AC}$, $\overline{BD}$ and $\overline{CD}$ respectively. Prove that $\triangle OPQ$ is equilateral. The condition $AD = BC,$ also is necessary for the andyciup’s problem. From cyclic Quadrilateral $CDEQ$ $\Longrightarrow$ $\angle ECQ = \angle EDQ$ $,(1)$ From $(1)$ $\Longrightarrow$ $\angle ADQ = \angle BCQ$ and so, the triangles $\bigtriangleup ADQ,$ $\bigtriangleup BCQ$ are congruent, because they have also $AD = BC$ and $DQ = CQ.$ Hence , we have that $AQ = BQ$ $,(2)$ and $\angle EAQ = \angle EBQ$ $,(3)$ From $(3)$ $\Longrightarrow$ the quadrilateral $ABQE$ is cyclic and hence, we have that $\angle AQB = \angle AEB = 60^{o}.$ That is the triangle $\bigtriangleup AQB,$ is equilateral and the proof is completed. Kostas Vittas.

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