We have a trapezoid $ABCD$ with base $AD$ larger than side $CD$. The bisector of $\angle D$ intersects side $AB$ at point $K$. Prove that $AK>KB$.
All that I have tried was to make such drawing in GeoGebra, which obviously showed me that $AK>KB$, even if I extend $AD$ very, very long. I think the solution should go somehow through similar triangles, but I honestly have no idea how. I would really appreciate any help you provide. To mention more, I seriously don't need the entire solution. Even a little hint would be very helpful for me, since I don't really know where to start.
EDIT: key mistake was made in the previous problem: it's not $AD$ that's larger than $BC$, but $AD$ is larger than side $CD$.