Question 52
In $\triangle ABC$ we have $\angle A = 62^\circ$ and $\angle B = 64^\circ$, and $D, E$ are the feet of the perpendiculars from $C, B$ to $AB, AC$ respectively. Suppose $P, Q$ are the intersection points of line $DE$ with the circumcircle of $\triangle ABC$, with $P$ on arc $\widehat{AB}$ and $Q$ on arc $\widehat{AC}$. Then $\angle QAE - \angle PAD = \underline{\qquad}^\circ$.