Since both AP, AQ are different from the bisector AC of the right angle $\angle DAB$, we have $BX < DX \le DY + XY,\ \ DY < BY \le BX + XY$ It remains to show that XY < BX + DY. Denote a = AB = BC = CD = DA the square side, 0 < p = BP = CQ < a, and $\beta = \angle PAB < 45^\circ,\ \delta = \angle QAD < 45^\circ$. $\tan \beta = \frac{BP}{AB} = \frac p a,\ \ \tan \delta = \frac{DQ}{DA} = \frac{a - p}{a}$ $\tan(\beta + \delta) = \frac{\tan \beta + \tan \delta }{1 - \tan \beta \tan \delta} = \frac{\frac p a + \frac{a - p}{a}}{1 - \frac{p(a - p)}{a^2}} = \frac{1}{1 - \frac{p(a - p)}{a^2}} > 1$ Consequently, $\beta + \delta > 45^\circ,\ 2\beta + 2\delta > 90^\circ.$ There are other ways to show this, for example, see the solution of problem Square, but it is not worth the effort. Let B', D' be reflections of the square vertices B, D in the lines AP, AQ, respectively. The quadrilateral BD'B'D is cyclic with the circumcenter A, because the perpendicular bisectors AP, AQ, AC of BB', DD', BD meet at the vertex A. Since $\angle BAB' = 2 \beta,\ \angle DAD' = 2\delta,$ we have $\angle BAB' + \angle DAD' > 90^\circ$ and consequently, the points B, D', B', D follow on the minor arc BD of the circumcircle (A) in this order. As a result, any circle (X) centered on the segment (AP) and with the radius BX = B'X contains the point D' in its interior and any circle (Y) centered on the segment (AQ) and with radius DY = DY' contains the point B' in its interior. Thus any 2 such circles (X), (Y) intersect, which means that $XY < B'X + D'Y = BX + DY$

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Let $ \omega_1$ and $ \omega_2$ be the circumferences centered at $ X,Y$ with radii $ XB$ and $ YD,$ respectively. It suffices to prove that $ \omega_1$ and $ \omega_2$ are secant. Let $ M \equiv BQ \cap AP$ and $ N \equiv DP \cap AQ.$ From $ \triangle ABP \cong \triangle BCQ$ and $ \triangle ADQ \cong \triangle DCP,$ it follows that $ AM \perp BQ$ and $ AN \perp DP$ $ \Longrightarrow$ $ T \equiv BQ \cap DP$ is the orthocenter of $ \triangle APQ.$ $ TP^2 - TQ^2 = AP^2 - AQ^2 = BP^2 - DQ^2 \Longrightarrow TX^2 - XB^2 = TY^2 - YD^2$ $ \Longrightarrow p(T,\omega_1) = p(T,\omega_2),$ which means that radical axis $ \zeta$ of $ \omega_1$ and $ \omega_2$ passes through $ T.$ Since $ BQ > BC = BA,$ then we have $ \angle BAQ > \angle AQB$ $ \Longrightarrow$ $ \angle DTQ > \angle TDQ$ $ \Longrightarrow$ $ QD > QT,$ $ YD > YT.$ Hence, $ T$ is always inside $ \omega_2$ $\Longrightarrow$ $ \zeta$ cuts $ \omega_2,$ say at $ U,V$ $ \Longrightarrow$ $ \omega_1$ and $ \omega_2$ meet at $ U,V$ and proof is completed.

Here I have a super beautiful solution by the argentinian Carlos di Fiore. Enjoy: Let $ PB=QC=a $ and $ BA=CP=b $ (obviously $ BA=AD=a+b $). Now consider in $ \mathbb{R}^3 $ the points $ A'=(a+b, 0, 0) $, $ P'=(0, a, 0) $ and $ Q'=(0, 0, b) $ (and $ O=(0,0,0) $). Since $ \angle PBA = 90^\circ = \angle P'OA' $, $ BA=a+b=OA' $ and $ PB=a=P'O $ we have that $ \triangle PBA \equiv \triangle P'OA' $. In the same way $ \triangle QDA \equiv \triangle Q'OA' $ and $ \triangle PCQ \equiv Q'OP' $. From that we have $ PA=P'A' $, $ QA=Q'A' $ and $ PQ=P'Q' $ so $ \triangle AQP \equiv \triangle A'Q'P' $. Now we consider $ X' $ and $ Y' $ in $ A'P' $ and $ A'Q' $ respectively such that $ AX=A'X' $ and $ AY=A'Y' $. Obviously $ \triangle AXB \equiv \triangle A'X'O' $, $ \triangle AYD \equiv \triangle A'Y'O $ and $ \triangle AXY \equiv \triangle A'X'Y' $ so $ BX=OX' $, $ DY=OY' $ and $ XY=X'Y' $. But it's obvious that $ \triangle X'OY' $ is a triangle, and has sides $ BX $, $ DY $ and $ XY $, hence the solution is complete $ \Box $.