This problem does not seem to make sense. How can a line be parallel to each of two lines (the diagonals) that are not parallel themselves?

It is obvious $EL \parallel FK$ and $FL \parallel EK$. It is well know if $EF$ is perpendicular bisector of $KL$ hence $ELFK$ is a rhombus and so we are done. Now we have to prove that $\angle KEF = \angle KFE$. Now suppose that $\angle ACB = 2\alpha , \angle CBD = 2 \beta$ So we have : \[ \angle KEF = \angle KEC + \angle CEF = \alpha + \beta = \angle BFE + \angle KFB = \angle KFE \]