Let $ ABCD$ be a trapezium of parallel sides $ AD$ and $ BC$ and non-parallel sides $ AB$ and $ CD$ Let $ I$ be the incenter of $ ABC$. It is known that exists a point $ Q \in AD$ with $ Q \neq A$ and $ Q \neq D$ such that if $ P$ is a point of the intersection of the bisectors of $ \widehat{ CQD}$ and $ \widehat{CAD}$ then $ PI \parallel AD$ Prove that $ PI=BQ$