Since $\angle PAQ=\angle BAC=45^{\circ}$ $\Longrightarrow$ $PQ=BC=CD$ $\Longrightarrow$ $BPCQ$ is an isosceles trapezoid $\Longrightarrow$ $PB=QC$ and similarly $PC=QD$ $\Longrightarrow$ $\triangle PBC \cong \triangle QCD$ $\Longrightarrow$ $PBDC \cong QCAD.$ Hence if $L \equiv PD \cap BC,$ it follows that $QN:QA=PL:LD=PM:MA$ $\Longrightarrow$ $MN \parallel PQ.$

My solution: $S$ is the projection of $A$ on $MN$ $\Rightarrow$ $AB = AS = AD$ (familiar property) It's easy to see that: $\triangle{APC}$$\sim$$\triangle{ASN}, \triangle{AQC}$$\sim$$\triangle{ASM}$ $\Rightarrow$ $\frac{AP}{AC} = \frac{AS}{AN}, \frac{AC}{AQ} = \frac{AM}{AS}$ $\Rightarrow$ $\frac{AP}{AQ} = \frac{AM}{AN}$ and the conclusion follows

Let $O$ be the centre of the square, $K=AM\cap BD, L=AN\cap BD$; $\angle MAL=\angle BML=45^\circ\implies ML\bot AN$. Similarly $NK\bot AM$, hence $MNLK$ is cyclic, wherefrom $\angle AMN=\angle ALK\ (\ 1\ )$. $PC\bot AP\implies PCOK$ cyclic, so $AK\cdot AP=AO\cdot AC\ (\ 2\ )$. Similarly $LOCQ$ is cyclic, therefore $AL\cdot AQ=AO\cdot AC\ (\ 3\ )$. From $(2)$ and $(3)$ $PQLK$ is cyclic, so $\angle KPQ=\angle ALK\ (\ 4\ )$; with $(1)$ and $(4)$ we are done. However Luis's solution is really original and effective! Best regards, sunken rock

Dear Mathlinkers, 1. MLN is the Pascal’s line of APDCBQA 2. PCQD is a trapeze and PD perpendicular to AQ 3. PBQC is a trapeze and BQ perpendicular to AP 4. L is the orthocenter of APQ 5. X the symmetric of L wrt PQ 6. ML is the Pascal’s line of XCBQPAX and ML // PQ and we are done without any calculation… Sincerely Jean-Louis

@sunken rock, how is ML perpendicular to AN? (how is angle AML equal to 45 degrees?)

A different approach: Let $AM$ cuts $BD$ at $M'$ and $AN$ cuts $BD$ at $N'$ it 's obvious that $DNM'A$ and $BMN'A$ are cyclic so $\angle M'NN'=\angle M'NA = \angle M'DA=\angle BDA , \angle N'MM'=\angle N'MA \angle N'BA=\angle DBA$ but $ABD$ is isoceles thus $MNM'N'$ besides we know $M'N'PQ$ is cyclic hence Reim's yields $MN \parallel PQ$ RH HAS differrent angle chase

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Let $O$ be the center of the circle. Clearly, $POQ=90^{\circ}$. Let $\angle DAN = 90^{\circ}$. Some angle chasing yields $\angle ANM=(x+45)^{\circ}$, so the problem reduces to showing $\angle AQP=(x+45)^{\circ}$. This is equivalent to showing $\angle AQO=x^{\circ}$, since $\angle OQP = 45^{\circ}$. Let $CD$ intersect $OQ$ at $X$. By $AA$, $NXQ$ is similar to $NDA$, which means $\angle AQO = x^{\circ}$, and the result follows.