Prove that $NK$ is tangent to the circumcircle of $\Delta KEF$ .

Question

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.

Answer 1

First, I recommend GeoGebra for geometry drawings. You would get much better images and more flexibility.

For the solution, we first observe that

1. $AP$ and $BQ$ are corresponding medians of two similar triangles $\triangle AKD$ and $\triangle BKC$, so $\angle PAK = \angle QBK$
2. $APQB$ is circular from the above statement.
3. $\triangle NPQ$ 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.