Notice $PQ = RS$ implies $N$ is also the midpoint of $PS$, then as $APS$ and $CQR$ are right angled it follows that $$\angle ANC = \angle ANP + \angle CNQ = 2(\angle ASN + \angle CRN) = 2(\angle DSR + \angle DRS) = 2\angle ADC$$ As $M$ is the center of cyclic queadrilateral $ABCD$ it follows that $\angle AMC = 2\angle ADC$ too, done.

Notice that $\angle MAN=\angle MCN \Leftrightarrow \angle NAS - \angle MAD=\angle MCD- \angle NCR$, and since $N$ is the midpoint of $PS$, we get that the later is equivalent with $\angle MDA+\angle CDM=\angle DRS +\angle RSD$, which is true as $\angle RDA$ is an exterior angle of $\Delta RDS$. Finally, we can conclude that $\angle MAN=\angle MCN$, hence $AMNC$ is a cyclic quadrilateral.

İs it really EGMO P1?too easy. The KEY is that in right angled triangle the median is equal to the half of hipotenous.

The condition is equivalent to $N$ being the midpoint of both $\overline{PS}$ and $\overline{QR}$ simultaneously. (Thus triangles $BAD$ and $BCD$ play morally dual roles.) The rest is angle chasing. We have \begin{align*} \measuredangle ANC &= \measuredangle ANP + \measuredangle QNC \\ &= 2\measuredangle ASP + 2\measuredangle QRC \\ &= 2 \measuredangle DSR + 2 \measuredangle DRS = 2 \measuredangle RDS \\ &= 2 \measuredangle ADC = \measuredangle AMC. \end{align*}

@juckter I believe you mean to type $PS$ not $PQ$.

Thanks, edited.

Let $W,X$ and $Y,Z$ be the feet of the altitudes from $A,C$ to $PQ$ and $BD$, respectively. Note that as $CQR$ and $ABD$ are right triangles, $\angle WAY$ is the reflection of $\angle NAM$ over the angle bisector of $\angle BAD$. Similarly, $\angle XCZ$ the reflection of $\angle NCM$ over the angle bisector of $\angle BCD$. As $AW,CX$ and $AY,CZ$ are pairs of parallel lines, $\angle WAY=\angle XCZ$, so $\angle NAM=\angle NCM$ as desired.

Notice that $\odot(APS)$ and $\odot(CRQ)$ have common circumcenter at $N$, so $$\angle ANQ+\angle CNQ=2\pi-2\angle APN-2\angle CQN=2\angle BPQ=2\angle ADC=\angle AMC$$as desired.

Extension: Prove that $(CQR), (ABCD), AN$ concur. (And similarly, $(APS), (ABCD), CN$ concur.)

What is the point of $\angle ABC > \angle CDA$? Configuration issues?

There is no circle if ABC equals CDA.

IstekOlympiadTeam wrote: Let $ABCD$ be a convex quadrilateral with $\angle DAB=\angle BCD=90^{\circ}$ and $\angle ABC> \angle CDA$. Let $Q$ and $R$ be points on segments $BC$ and $CD$, respectively, such that line $QR$ intersects lines $AB$ and $AD$ at points $P$ and $S$, respectively. It is given that $PQ=RS$.Let the midpoint of $BD$ be $M$ and the midpoint of $QR$ be $N$.Prove that the points $M,N,A$ and $C$ lie on a circle. Here is my solution. Because $PQ=RS$ the point $N$ is the midpoint of $PS.$ Because $\angle SAP=\angle QCR=90^\circ,$ we deduce that the point $N$ is the circumcenter of both triangles $SAP$ and $QCR.$ Therefore, \begin{align*}\angle CNA &=\angle CNQ+\angle PNA\\&=2\angle CRQ+2\angle PSA \\&= 2(\angle DRS+\angle RSD) \\&= 2\angle CDA=\angle CMA \end{align*}and this proves that the points $M,N,A$ and $C$ lie on a circle.

I understand that the solution is just easy angle chasing, but just for the sake of curiosity, is it "legal" to make projective transformation that sends $AC \cap BD$ to the center of circumcircle $ABCD$ or anything similar? If then, does the new line $A'M'N'C'$ (they become collinear in the new configuration) map to a circle in the original configuration?

Yeah but you need to show that $M,N,A,C$ are concyclic which isn't something projectivity saves .(for example a perspectivity).

Solution(with char_2539) We see that $M$ is the circumcenter of cyclic quadrilateral $ABCD \Longrightarrow MA = MB = MC = MD$ N is the circumcenter of $\triangle RCQ$ and $\triangle ASP \Longrightarrow NC = NQ = NR$ ; $NA = NP = NS$ $\angle DBA = \alpha $ , $\angle DBC = \beta$ and $BQP = \gamma$ See that $\angle BCA = \angle DAM = 90 - \alpha$ and $\angle CAB = 90 - \beta$ and $\angle NCQ = \gamma$ Also, $\angle MCA = \angle MAC = \alpha + \beta - 90 \Longrightarrow \angle NCM = \beta - \gamma$ $\angle APS = \alpha + \beta - \gamma \Longrightarrow \angle ASP = \angle NAS = 90 - \alpha - \beta + \gamma$ Thus, $\angle MAN = \beta - \gamma$ $MNCA$ is cyclic.

Since $N$ is the center of both $(CQR)$ and $(APS)$, we have that $$\angle MCN = \angle MCD - \angle NCR = \angle BDR - \angle DRS = \angle DSR - \angle ADM = \angle DAN - \angle DAM = \angle MAN$$, as desired.

The key observation here is that $M$ is the center of $(BAD)$ and $N$ is the center of $(APS)$ and $(QCR)$. We can proceed with angle-chase. Let $\angle ABD = a, \angle PSA = b, $ and $\angle BDR = c$. Then, we have that $\angle MDA = 90-a, \angle QBD = 90-c, $ and $\angle APS = 90-b$. Clearly then, looking at $\triangle PQB$, $\angle PQB = b+a-c$. Therefore, since $\angle QCN = b+a-c$ and $\angle QCM = 90-c$, $\angle MCN = b+a-90$. On the other hand, $\angle PAN = 90-b$ and $\angle BAM = a$, so $\angle NAM = b + a -90$. Therefore, $\angle MCN = \angle NAM$ and $MNAC$ is cyclic.

From the condition that $RS=PQ$, we have that $NS=NP$, thus we have that $N$ is the center of $\left(APS\right)$, also we have that $N$ is the center of $\left(CRQ\right)$. We denote with $x=\angle DRS=\angle PRC$, and we denote with $\beta =\angle ADC$.Then we have the following: $$\angle ANC=\angle ANP+\angle PNC=2.(\angle ASN+\angle PRC)=2.(180-(180-\beta+x)+x)=2\beta=\angle AMC$$thus we have that $AMNC$ is cyclic

We have $$\measuredangle AMC = 2\measuredangle ADC = 2\measuredangle DSR + 2\measuredangle SRD = \measuredangle ANP + \measuredangle PNC = \measuredangle ANC$$ as desired.

Notice $PQ=RS$ implies $N$ is the midpoint of $PS,$ so $N$ is the center of $(CQR)$ and $(APS),$ while $M$ is the center of $(ABC).$ Notice $$\angle BDA+\angle CDB=\angle CDA=\angle RSD+\angle DRS,$$so \begin{align*}\angle MAN&=\angle DAN-\angle DAM=\angle NSA-\angle MDA\\&=\angle CDM-\angle DRS=\angle MCR-\angle NCR=\angle MCN.\end{align*}$\square$

Probably same as the above solutions but still posting. Solved with Krutarth and Malay. Note that $N$ is the midpoint of $PS$. And since $\angle DCB=90\implies NC=NQ=NR.$ Similarly, we have $AN=AP=AS$. So we have $$ \angle ANC=\angle ANP+\angle PNC$$$$=180-\angle 2\angle APN+180-2\angle BQP=360-2(\angle APN+ \angle BQP)=2(180-(\angle APN+ \angle BQP))$$$$=2\angle CBP=2\angle ADC.$$ But since $M$ is the center of $ABCD$ we have $$\angle ANC= 2\angle ADC=\angle AMC\implies ANMC\text{ cyclic}.$$

We will show that $\angle CMA = \angle CNA$, which is enough to imply the result. This is true because $$\measuredangle CNA = \measuredangle CNQ + \measuredangle QNA = 2\measuredangle CQR + 2\measuredangle NPA = 2\measuredangle QBP = \measuredangle CMA$$by utilizing that $M$ is the center of the circle and $N$ is a common midpoint.

Note that $N$ is the center of $(APS)$ and $M$ is the center of $(ABCD)$. Now \[ 2 \angle D = \angle CMA \]and \[ \angle CNA= \angle CNQ +\angle ANQ = (180-2 \angle PQB )+(180 - 2 \angle NAP) = 360 - 2(\angle NPB+\angle PQB) = 2\angle D\]

Let $\angle ADC=\alpha$ and $\angle ASC=\beta.$ THen, $\angle AMC=2\alpha.$ However, since $N$ is the midpoint of $PS$, $\angle ANP=2\beta.$ Then, $$\angle PNC=2\angle NKC=2(\alpha-\beta),$$so $\angle ANC=\angle ANP=\angle PNC=2\alpha,$ hence done.

Easiest of all EGMO geometry. $$\angle{AMC}=2\angle{ADQ}$$. $$\angle{ANC}=\angle{ANQ} + \angle{PNC}=2\angle{ASN}+2\angle{NRC}=2\angle{ANC}+ 2\angle{DRS}=2\angle{ADQ}$$. Hence this proves that $MNCA$ is a cyclic quadrilateral.

It's simply angle chasing. Mine is not ideal but here it is. $ABCD$ is cyclic $\implies$ $\angle{ADC}=\angle{PBQ}$ From median in a right triangle we have: $\angle{ABM}=\angle{BAM}=90^{\circ}-\angle{ADB}$ $\angle{BCM}=\angle{CBM}=90^{\circ}-\angle{BDC}$ $\angle{APN}=\angle{PAN}$ $\angle{CQN}=\angle{QCN}$ $\angle{MCN}=\angle{QCN}-\angle{QCM}=\angle{CQN}-(90^{\circ}-\angle{BCD})=\angle{PQB}+\angle{BDC}-90^{\circ}=180-\angle{PBQ}-\angle{BPQ}+\angle{BDC}-90^{\circ}=90-\angle{ADB}-\angle{APN}=\angle{BAM}-\angle{BAN}=\angle{MAN}$ $\implies AMNC$ is cyclic qed.

This is a cringe solution Note that $CN$, $CM$, $AM$ and $AN$ are medians in right angled triangles Let $\angle CDB = \alpha$ and $\angle ADB = \beta$ and $\angle MCN = \psi$. $\angle MDC = \angle DCM = \alpha$ and $\angle DCM = \alpha - \psi$ and $\angle MDA = \angle MAD = \beta$. From $\triangle CRQ \implies \angle CRN = \alpha - \psi$. Then $\angle DRS = \angle NRC = \alpha - \psi \implies \angle DSR = \beta + \psi$ Since $AN$ is a median in the right angle $\triangle SAP \implies \angle SAN = \angle ASN = \beta + \psi$, but $\angle MAD = \beta \implies \angle MAN = \psi$. Combining with $\angle MCN = \psi \implies A, M, N C$ are concyclic.

We can obviously see that $M$ is the curcumcenter of $ABCD$. We can also see that $N$ is the midpoint of $PS$ and $QR$ which creates a lot of isosceles triangles. We claim that $\angle AMC = \angle ANC = 2 \cdot \angle ADC $. We prove this for $\angle AMC$ first. We can see that \[\angle AMC = \angle AMB + \angle BMC \Rightarrow \angle MAB + \angle MCB = 180 - \frac{\angle AMC}{2} \Rightarrow \angle ADC = \angle DAM + \angle DCM = \frac{\angle AMC}{2} \]For $\angle ANC$ we have \[ \angle ANC = \angle ANP + \angle NQC \Rightarrow \angle NAP + \angle NCQ = 180 - \frac{\angle ANC}{2} \Rightarrow \angle DAN + \angle ANC + \angle NCD + \angle ADC = 360\](note that this is not convex). Thus we can see that, $\frac{\angle ANC}{2} + 360 - \angle ANC + \angle ADC = 360$. $\blacksquare$

The condition implies $N$ is the midpoint of $\overline{PS}$. Now, $$\angle ANC=\angle ANP+\angle CNQ=2(\angle ASP+\angle CRQ)=2(\angle DPR+\angle DRP)=2\angle ADC=\angle AMC,$$since $M$ is the cicrumcenter of $ABCD$. $\blacksquare$

Notice that $N$ is the midpoint of $\overline{SP}$ as well. Thus, we have \[\angle ANC = \angle ANP + \angle QNC = 2(\angle ASP + \angle QRC) = 2(\angle DSR+\angle DRS) = 2 \angle ADC\] Then, since $M$ is the center of cyclic quadrilateral $ABCD$, we have $\angle AMC = 2 \angle ADC$. This means that $\angle AMC = \angle ANC$, so $AMNC$ is cyclic. $\square$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -3.574715947923021, xmax = 10.84089483568482, ymin = -2.17273844548352, ymax = 6.725248240918899; /* image dimensions */ /* draw figures */ draw((0,4)--(0,0), linewidth(1)); draw((0,0)--(3,0), linewidth(1)); draw((3,0)--(3.996502867506914,2.1321871118141122), linewidth(1)); draw((3.996502867506914,2.1321871118141122)--(0,4), linewidth(1)); draw((0,4)--(0,4.647781263843776), linewidth(1)); draw((0,4.647781263843776)--(5.031359934663131,0), linewidth(1)); draw((5.031359934663131,0)--(3,0), linewidth(1)); draw((0,4)--(3,0), linewidth(1)); draw((1.5,2)--(2.5156799673315655,2.323890631921888), linewidth(1)); draw((2.5156799673315655,2.323890631921888)--(3.996502867506914,2.1321871118141122), linewidth(1)); draw((1.5,2)--(0,0), linewidth(1)); draw((0,0)--(3.996502867506914,2.1321871118141122), linewidth(1)); draw((0,0)--(2.5156799673315655,2.323890631921888), linewidth(1)); draw((1.5,2)--(3.996502867506914,2.1321871118141122), linewidth(1)); /* dots and labels */ dot((0,0),dotstyle); label("$A$", (0.0054334115247282,-0.15628020336219746), NE * labelscalefactor); dot((3,0),dotstyle); label("$B$", (3.04425208963368,0.03628020336219746), NE * labelscalefactor); dot((0,4),dotstyle); label("$D$", (0.03554334115247282,4.008193333153482), NE * labelscalefactor); dot((3.996502867506914,2.1321871118141122),dotstyle); label("$C$", (4.036552075494734,2.1435417113706215), NE * labelscalefactor); dot((3.6125462655944176,1.3106467582555346),dotstyle); label("$Q$", (3.654898234778944,1.314807657244906), NE * labelscalefactor); dot((1.419326061944203,3.336661176302054),dotstyle); label("$R$", (1.413114749525407,3.4030252107631043), NE * labelscalefactor); dot((5.031359934663131,0),linewidth(4pt) + dotstyle); label("$P$", (5.032469643151878,0.05447141246415232), NE * labelscalefactor); dot((0,4.647781263843776),linewidth(4pt) + dotstyle); label("$S$", (0.04554334115247282,4.6297438737477675), NE * labelscalefactor); dot((1.5,2),linewidth(4pt) + dotstyle); label("$M$", (1.439445517668565,2.0908801750843056), NE * labelscalefactor); dot((2.5156799673315655,2.323890631921888),linewidth(4pt) + dotstyle); label("$N$", (2.514458689876687,2.40710764310596), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Angles Note that $$\angle ANC=360^{\circ}- \angle CNR- \angle ANR =180^{\circ}- \angle CNR+180^{\circ}- \angle ANR $$$$=2\angle ASN+ 2\angle RNC= 2\angle DSR+2\angle DRS= 2(180^{\circ}-\angle RDS)=2\angle ADC= \angle AMC$$ Which means $A$,$M$, $N$, and $C$ are cyclic.

$PQ = RS$, $NQ=NR\implies N$ is the midpoint of $PS$. Now, \begin{align*} \measuredangle NCM &=\measuredangle NCQ-\measuredangle MCB\\ &= (90^\circ - \measuredangle NRC) - (90^\circ - \measuredangle BDC)\\ &=\measuredangle BDC - \measuredangle SRD\\ &=(\measuredangle BDA+\measuredangle SDC) - \measuredangle SRD\\ &=\measuredangle BDA + \measuredangle SPR + \measuredangle DRS\\ &=\measuredangle BDA + \measuredangle DSR\\ &=\measuredangle BDA -\measuredangle RSD .\end{align*} And, \begin{align*} \measuredangle NAM &= \measuredangle BAM - \measuredangle BAN\\ &=-(90^\circ - \measuredangle BDA)-\measuredangle PAN\\ &=\measuredangle BDA - 90^\circ-(-(90^\circ-\measuredangle PSA))\\ &=\measuredangle BDA -90^\circ + 90^\circ - \measuredangle PSA\\ &=\measuredangle BDA \measuredangle PSA\\ &=\measuredangle BDA - \measuredangle RSD .\end{align*} Thus $\measuredangle NCM=\measuredangle NAM \implies MNAC$ is cyclic.

We are given that $\square{ABCD}$ is cyclic with center $M$ and $N$ is the midpoint of both $PS$ and $QR$, the respective hypotenuses of right triangles $\Delta{APS}$ and $\Delta{CQR}$. Hence, we have: $\measuredangle{ANC} = \measuredangle{ANP} + \measuredangle{QNC} = 2 \measuredangle{NAP} + 2 \measuredangle{QCN} = 2 \measuredangle{NAB} + 2 \measuredangle{BCN} = 2 (\measuredangle{ANC} + \measuredangle{CBA}) = 2 \measuredangle{ABC} = 2 \measuredangle{ADC} = \measuredangle{AMC}$.

$N$ is the circumcenter of $(CRQ)$ and $(ASP).$ So \[\angle ANC=\angle CNQ+\angle ANP=2(\angle CRQ+\angle ASP)=2(\angle DRS+\angle DSR)=2\angle ADC=\angle AMC.\]

Pretty similar to the above solutions, but this is for storage. Notice that $M$ is the circumcenter of both $(ABD), (CBD),$ while $N$ is the circumcenter of both $(APS), (CQR).$ Therefore, $$\angle ANC = \angle CNQ + \angle ANP = 2(\angle ASP+\angle CRQ) = 2\angle ADR = 2(\angle ADB+\angle BDC) = \angle AMB+\angle BMC = \angle AMC,$$and the desired result follows. QED