We have $\mathrm{DY}\angle \mathrm{DAY}$ ). Similarly, $\mathrm{BX}\angle \mathrm{PAQ}$, so it follows by the same observation as above that $\mathrm{XY}^{\prime}>\mathrm{XY}$. But the claim is almost obvious. Note that $\mathrm{PQ}^{\prime}=\mathrm{AB}$

So take $\mathrm{P}^{\prime}$ on $\mathrm{AD}$ with $\angle \mathrm{P}^{\prime} \mathrm{PQ} \mathrm{P}^{\prime}=90^{\circ}$. Then $\mathrm{A}$ lies inside the circle $\mathrm{P}^{\prime} \mathrm{PQ}^{\prime}$, so extend $\mathrm{PA}$ to meet it again at $\mathrm{A}^{\prime}$. Then $\angle \mathrm{PA}^{\prime} \mathrm{Q}^{\prime}=\angle \mathrm{PP}^{\prime} \mathrm{Q}^{\prime}=45^{\circ}$, so $\angle \mathrm{PAQ}^{\prime}=\angle \mathrm{PA}^{\prime} \mathrm{Q}^{\prime}+\angle \mathrm{AQ}^{\prime} \mathrm{Q}^{\prime}>45^{\circ}$. But $\angle \mathrm{PAQ}^{\prime}+\angle \mathrm{PAQ}=90^{\circ}$, so $\angle \mathrm{PAQ}^{\prime}>\angle \mathrm{PAQ}$ as claimed.