Prove that $NK$ is tangent to the circumcircle of $\Delta KEF$ . Consider a circle $O$ with radius $R$. $ABCD$ is cyclic quadrilateral, the intersection of $AC$ and $BD$ is $K$. $P$ and $Q$ are respectively the midpoints of   $KD$ and $KC$. The intersection of $AP$ and $BQ$ is $N$, $M$ is midpoint of $CD$, the intersection of $MN$ and $AC$ is $E$, $MN$ and $BD$ is $F$. Prove that $NK$ is tangent to the circumcircle of $\Delta KEF$ .

I think this is hard problem and i did not know where to start the idea for solve this problem. Help me, thanks.
 A: First, I recommend GeoGebra for geometry drawings.  You would get much better images and more flexibility.

For the solution, we first observe that 


*

*$AP$ and $BQ$ are corresponding medians of two similar triangles $\triangle AKD$ and $\triangle BKC$, so $\angle PAK = \angle QBK$

*$APQB$ is circular from the above statement.

*$\triangle NPQ$ and $\triangle NBA$ are similar, from (2).


Now, the fact that $NK$ is tangent to circle $KEF$ is equivalent to $\angle EKN = \angle KFE$, and that what we will prove here. 
Since $MQ\parallel BD$, $\angle KFE = \angle EMQ$. So we need only prove $\angle AKN = \angle QMN$. But then, that is done if we are able to show 
$$\triangle AKN \text{ and } \triangle QMN\text{ are similar.}\tag{4}$$
Note that 
$$\angle MQN = \angle FBN = \angle KAN,$$
where the first equality comes from $MQ\parallel FB$ and the second comes from (3) above.
So, in order to show (4), we only need
$$\frac{MQ}{AK} = \frac{NQ}{NA}.$$
But from (3), we see that
$$\frac{NQ}{NA} = \frac{PQ}{AB} = \frac{CD}{2 AB}.\tag{5}$$
Since $\triangle ABK$ and $\triangle DCK$ are similar
$$\frac{CD}{AB} = \frac{DK}{AK}.\tag{6}$$
From (5) and (6), and the fact that $DK = 2MQ$, we have
$$\frac{NQ}{NA} = \frac{DK}{2 AK} = \frac{MQ}{AK}.$$
And we are done.
