socrates wrote: Let $ABCD$ be a parallelogram such that $\angle BAD = 60^{\circ}.$ Let $K$ and $L$ be the midpoints of $BC$ and $CD,$ respectively. Assuming that $ABKL$ is a cyclic quadrilateral, find $\angle ABD.$ Let $AB, LK$ intersect at $E$, $EB=a,AB=2a$ and $EK=KL=b, BD=2b$. I draw $BH$ perpendicular to $AB$. It is: $EB \cdot EA = EK \cdot EL \Leftrightarrow 3{a^2} = 2{b^2} \Leftrightarrow b = \frac{{a\sqrt 6 }}{2}$ We also have, $BH = a\sqrt 3 \Leftrightarrow BD = 2b = a\sqrt 3 \sqrt 2 = BH\sqrt 2 $, thus $\omega=45^0$ and finally $\boxed{\angle ABD=75^0}$

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Yes that's exactly my solution! Easy problem