Show that $MNPQ$ is a square Let $ ABCD $ a quadrilateral s.t. $AC=BD $ and $m (\angle AOD)=30°$ where $O=AC\cap BD $.
Let $\triangle ABM, \triangle DCN, \triangle ADN, \triangle CBQ $ equilateral triangles with $Int (\triangle ABM)\cap Int (ABCD)=\emptyset$,  $Int (\triangle DCP)\cap Int (ABCD)=\emptyset$,  $Int (\triangle ADN)\cap Int (ABCD) \neq \emptyset$,  $Int (\triangle CBQ)\cap Int (ABCD)\neq\emptyset$.
Show that $MNPQ$ is a square.
I have no idea how to start.

Let $R^{\alpha}_O$ be a rotation in the plain by an angle $\alpha$ around a point $O$.
Easy to see that to rotate a vector by an angle $\alpha$ it's the same to rotate this vector around his tail.
Now, by using the beautiful  Daniel Mathias's picture we obtain:
$$R^{90^{\circ}}\left(\vec{NM}\right)=R^{30^{\circ}}\left(R^{60^{\circ}}\left(\vec{NA}+\vec{AM}\right)\right)=R^{30^{\circ}}\left(\vec{DA}+\vec{AB}\right)=$$
$$=R^{30^{\circ}}\left(\vec{DB}\right)=R^{60^{\circ}}\left(\vec{AC}\right)=R^{60^{\circ}}\left(\vec{AB}+\vec{BC}\right)=\vec{MB}+\vec{BQ}=\vec{MQ},$$
which says $NM\perp MQ$ and $NM=MQ.$
Also,
$$R^{90^{\circ}}\left(\vec{QP}\right)=R^{30^{\circ}}\left(R^{60^{\circ}}\left(\vec{QC}+\vec{CP}\right)\right)=R^{30^{\circ}}\left(\vec{BC}+\vec{CD}\right)=$$
$$=R^{30^{\circ}}\left(\vec{BD}\right)=R^{60^{\circ}}\left(\vec{CA}\right)=R^{60^{\circ}}\left(\vec{CB}+\vec{BA}\right)=\vec{QB}+\vec{BM}=\vec{QM},$$
which says $QM\perp PQ$ and $QM=PQ.$
Can you end it now?
