Here is the question:

In convex quadrilateral $ABCD$, let $M$ be the midpoint of $\overline{BC}$, and $\angle AMD=120^\circ$.  Show that $AB + \dfrac{1}{2}BC + CD \geq AD$.

When rotation works, usually there are some equal lengths, such as the sides of a square of an equilateral triangle.  In this case, the only equal lengths are $BM$ and $MC$, and it is not easy to see how a rotation from $BM$ to $MC$ helps with the inequality.  Can you consider reflection?  Also, it might not be triangle inequality because there are three terms on the left hand side; of course the principle that "the straight line is the shortest path between two given points" is still at play.