It is known that $∠BAD =\frac{π}{3}$ and that the points $A$, $B$, $K$, $L$ lie on a common circle. Find $\angle ADB$. Consider a parallelogram $ABCD$ and let $K$, $L$ be the midpoints of $BC$ and $CD$. It is known that $∠BAD =\frac{π}{3}$ and that the points $A$, $B$, $K$, $L$ lie on a common circle. Find $\angle ADB$.
I struggle greatly with geometry and I tried to use inscribed angles inside a circle and properties of parallelograms but I can't quite create the right picture to set up the correct assumptions. 
 A: 
Given the parallelogram ABCD and the midpoints L and K, we have LK || DB, AX : AB = 3:2 and  ∠ABC = 120, ∠DBC = ∠LKC = $x$, ∠LKB = 180 -$x$. 
Since ABKL is cyclic, $\angle$ALK = 180 -  ∠ABC = 60 and ∠LAB = 180 - ∠LKB = $x$.
Apply the sine rule to the triangles DAB and LAX,
$$\frac {\sin\angle ADB}{\sin \angle DAB}=\frac {\sin x}{\sin 60} = \frac{AB}{DB}, \>\>\>\>\>
\frac {\sin\angle LAX}{\sin \angle ALX}=\frac {\sin x}{\sin 60} = \frac{LX}{AX} $$
Recognize DB = LX, AB = $\frac23$AX and multiply the two equations to get,
$$\frac {\sin^2 x}{\sin^2 60} = \frac23\implies \sin x = \frac1{\sqrt2}\implies x =45$$
A: 
Please note that this post does not contain the answer to your question. Instead, I am giving you the sketch, which you were unable to draw to set up your assumptions. Ok, what the heck, I will give you the answer too. It is $\measuredangle ADB=\phi=45^0$. Now, it is up to you to prove it. 
A: Here are some more details on the path I proposed you. I invite you to fill in the missing details. 
As in YNK's drawing, call $\measuredangle ADB=\phi$. Let $H$ on $BD$ such that $AH\perp BD$ and $I = AL \cap BD$.



*

*Since $\square ABKL$ is cyclic, $\angle ALK$ and $\angle ABC$ are supplementary.

*Use Thales Theorem and angle chasing to show that $\measuredangle ALD = \phi$.

*Demonstrate that $\triangle AIB\sim \triangle DIL$, implying $\overline{AI} = 2\cdot \overline{IL}$.

*Demonstrate that $\triangle ADL \sim \triangle ADI$, so that $\overline{AD}^2 = \overline{AI}\cdot\overline{AL}$. 

*By means of 3. and 4. conclude that $\sqrt 3 \cdot \overline{AI} = \sqrt 2 \cdot \overline{AD}$.

*Recalling 1. and 2., show that condition 5. implies $\overline{AH} = \frac{\sqrt 2}2\cdot \overline{AD}$.

*Pythagorean Theorem on $\triangle ADH$ yields $AH \cong DH$, and thus $\phi = 45^\circ$.$\blacksquare$
