As shown in the figure (it's not very convenient to draw here, so draw it yourself),

Extend the extension lines of DG, DH and AB to m, Q.

∫AB∨CD

∴△BMG∽△CDG

∴CBDM=CBGG=2 1

Let AB=3a, then BM=23a.

And △AMK∽△EDK.

∴EAKK=ADEM=3×a23a=29①

Similarly, it is obtained by △BQH∽△CDH.

BQ∶CD=2∶ 1。

Because CD=AB=3a and BQ=6a.

And △AQN∽△FDN.

∴FANN=FADQ=3a2+a6a=29②

From ① and ②, we get EAKK=FANN.

∴KN∥EF

That is KN∑CD.