In the quadrilateral $ABCD$, $AD$ is parallel to $BC$. $K$ is a point on $AB$. Draw the line through $A$ parallel to $KC$ and the line through $B$ parallel to $KD$.
Prove that these two lines intersect at some point on $CD$.
Let $Q \equiv AB \cap CD$. Suppose that $A$ and $D$ are points on segments $BQ$ and $CQ$ respectively.
Let $L$ be the intersection of the parallel to $CK$ that goes through $A$ with $CD$. Easy to see $L$ is on segment $CD$.
Let $L'$ be the intersection of the parallel to $DK$ that goes trhough $B$ with $CD$. Easy to see that $L'$ is on segment $CD$.
We have that $\frac {QA}{QK}= \frac {QL}{QC} => QL \cdot QK = QA \cdot QC$.
Also $\frac{QB}{QK} = \frac{QL'}{QD} => QL' \cdot QK = QD \cdot QB$
But since $AD$ is parallel to $BC$ we have that $QA \cdot QC = QD \cdot QB$
Hence $QL \cdot QK = QL' \cdot QK => QL = QL' => L=L'$