elegant wrote: A circle with center O is inscribed in a quadrilateral ABCD and touches its non-parallel sides BC and AD at E and F respectively. The lines AO and DO meet the segment EF at K and N respectively, and the lines BK and CN meet at M. Prove that the points O,K,M and N lie on a circle. From Russian MO 1994. The basic idea the following well-known lemma, which is not hard to prove: In triangle $ABC$, suppose the incircle with center $I$ is tangent to $AB$ and $AC$ at $F$ and $E$ respectively. Let $BI$ meet $FE$ at $N$. Then $\angle BNC=90^{\circ}$. With the lemma, you can prove that $\angle ONM=\angle OKM=90^{\circ}$, which yields the concyclicity of $O,K,M,N$.

I understand lemma but... elaborate on it PLEASE.

elegant wrote: I understand lemma but... elaborate on it PLEASE. So where are you confused?

Why ∠ONM＝∠OKM＝90°

elegant wrote: Why ∠ONM＝∠OKM＝90° It follows directly from the lemma If $AD,BC$ intersect at $Z$. Apply the lemma to triangle $ZDC$, with $DO$ the angle bisector

Also, you should try to think by yourself, rather than asking people for everything. You cannot expect other people to help you in every situation. The real world is more cruel than you think

And one more thing: You should learn to use $\LaTeX$. It will make everything so much easier to read