Let $\angle MBC=\angle MCB=\alpha$. Then, $\angle MKD=\angle MND=\angle MCB=\alpha$. Let $\angle NDM=\angle KDM=\beta$. Since $DM\perp NK$, we find that $\angle DKN=90^\circ-\beta$. Hence, $\angle MNK=\angle MKN=90^\circ-\alpha-\beta$. Since $|CM|=|AB|=|CD|$, we find that $\angle CDM=\angle CMD=\angle MND+\angle MDN=\alpha+\beta\Rightarrow \angle CDK=\alpha$. Then, $\angle ABC=\angle ADC=2\beta+\alpha\Rightarrow ABM=2\beta\Rightarrow \angle MAB=90^\circ-\beta$. Also, $\angle DAB=180^\circ-\angle ABC=180^\circ-2\beta-\alpha\Rightarrow \angle DAM=90^\circ-\alpha-\beta$. Since $\angle NAM=90^\circ-\alpha-\beta=\angle NKM$, we have $ANMK$ is cyclic. Then, $\angle MAK=\angle MNK=\angle MKN=\angle MAN\Rightarrow \dfrac{|KA|}{|AL|}=\frac{|KM|}{|ML|}$....(1) Also, $\angle LDM=\angle KDM$. Hence, $\dfrac{|KD|}{|DL|}=\frac{|KM|}{|ML|}$....(2) Now, look at the locus of the points $X$ such that $\dfrac{|KX|}{|XL|}=\frac{|KM|}{|ML|}$. We know that these points lie on a circle. By (1) and (2), this circle is $(ADM)$. Since $P\in (ADM)$, we find that $\dfrac{|KP|}{|PL|}=\frac{|KM|}{|ML|}\Rightarrow \angle KPM=\angle LPM$. Then, $$\angle DPL=\angle MPL-\angle MPD=\angle MPK-\angle MAD=\angle MPK-\angle CNK=\angle MPK-\angle CPK=\angle CPM\text{ , as desired.}$$

a somewhat different solution: call circles $AMD,CMN$ as $\omega_1,\omega_2$, and their second intersection as $Q$.Translate $M$ with vector $\overrightarrow{BA}$ and call it $O$. it is clear that $DOMC, OMBA$ are rhombi. also see that $\widehat{LMD}=\widehat{CMD}=\widehat{CDM}$ so $LM \| DC$ so $LMKO$ are collinear. Now by a straightforward angle chase it is clear that $2\widehat{DMA}+\widehat{AOD}=2\widehat{DMA}+\widehat{CMD}=360$ so $O$ is the center of $\omega_1$. We know that $O,C$ are reflections of eachother in $DM$ and so are $N,K$ so this gives us that $ONKC$ is cyclic hence $O \in \omega_2$. now we prove that $L$ lies on the radical axis of $\omega_1,\omega_2$, it is equivalent to showing $OM^2-OL^2=OL\times OK$. using $OL=OM-LM$, $OK=ML+MN$, $MN=MK$ and $MC=MO$, this is equivalent to Thales in $NDC,NLM$ which is correct. so $P,L,Q$ are collinear. now note that $OP=OQ$ so $OC$ is the bisector of $\widehat{PCQ}$ and is the perpendicular bisector of $DM$. if we call the second intersection of $PC,\omega_1$ $R$, we see that $DM \| QR$ because $OC$ is perpendicular to both (note that $ORC, OQC$ are similar) so finally, $$\widehat{CPM}=180-\widehat{MPR}=\widehat{MQR}=\widehat{DRQ}=\widehat{DPQ}=\widehat{DPL}$$so we are done

A nice configuration with lots to explore. This was the slickest solution I found. Let $F$ be center of $(MAD)$. Note $180^\circ - \angle MDC = 180^\circ - \angle DMC = \angle DMN = \angle DMK$ so $MK \parallel AB,CD.$ Note \begin{align*} 2\angle AMB + 2\angle DMC + \angle BMC &= 180^\circ - \angle DCM + 180^\circ - \angle MBA + 180^\circ - \angle MCB - \angle CBM\\ &= 540^\circ - \angle DCB - \angle CBA \\ &=360^\circ \end{align*}so the reflections of $B$ over $AM$ and $DM$ coincide. This point is clearly $F.$ So $FM \parallel AB, CD$ and lines up with $LK.$ Note $\angle LDM = \angle MDK,$ and $LK$ passes through $F,$ so $(MAD)$ is the Apollonius circle of $L,K.$ Since $P$ lies on $(MAD),$ $\angle MPL = \angle MPK$ and it suffices to show $\angle MPD = \angle CPK.$ But $\angle MPD = \frac{1}{2} \angle DFM = \angle CFK = \angle CPK$, the end. $\blacksquare$

International Zhautykov Olympiad 2022/3 wrote: In parallelogram $ABCD$ with acute angle $A$ a point $N$ is chosen on the segment $AD$, and a point $M$ on the segment $CN$ so that $AB = BM = CM.$ Point $K$ is the reflection of $N$ in line $MD$. The line $MK$ meets the segment $AD$ at point $L$. Let $P$ be the common point of the circumcircles of $AMD$ and $CNK$ such that $A$ and $P$ share the same side of the line $MK$. Prove that $\angle CPM = \angle DPL$. Definitely not as clever as above's solution. This is a solution where I forgot Apollonian circles exist. Let $E$ be the center of $(AMD)$. Claim 01. $KL \parallel CD$. Proof. We have $\measuredangle KMD = \measuredangle NMD = \measuredangle CMD = \measuredangle MDC$. Claim 02. $K,L,M,E$ is collinear. Proof. It suffices to prove that $ME \parallel CD$. To see this, $\angle EMD = 90^{\circ} - \angle MAD = \angle MDC$. Claim 03. $ENKC$ is cyclic. Proof. Notice that $K$ is the reflection of $N$ wrt $MD$ by definition. We also have $EM = ED$, $CM = CD$ and $EM \parallel CD$, which forces $C$ to be the reflection of $E$ wrt $MD$ as well. Therefore, $NK$ and $EC$ shares the same perpendicular bisector, i.e. $NKEC$ is an isosceles trapezoid which forces $NKCE$ to be cyclic. Claim 04. $A,D,E,K$ is cyclic. Proof. We have $\angle KDA = \angle KDN = 2 \angle MDN = 2 \angle MDA = \angle MEA = \angle KEA$. Claim 05. $\angle MPL = \angle KPM$. Proof. Let the antipode of $M$ at $(AMD)$ be $M'$. We claim that $(K,L;M,M') = -1$. Here's an approach when I was high: Indeed, it suffices to prove that $KM \cdot KM' = KL \cdot KE$. Radical Center Theorem on $(AMD), (NPC), (ADK)$ gives us $AD \cap EK = L$ lies on $PQ$, i.e. $\text{Pow}_L(AMD) = \text{Pow}_L(NPC)$, which means $LM \cdot LM' = LK \cdot LE$. Therefore, \begin{align*} KM \cdot KM' &= (KL - LM)(KL + LM') = KL^2 - KL \cdot LM + KL \cdot LM' - LM \cdot LM' \\ &= KL^2 - KL \cdot LM + KL \cdot LM' - LK \cdot LE \\ &= KL(KL - LM + LM' - LE) = KL \cdot KE \end{align*}which gives what we wanted. Alternative Proof. Note that $MD$ bisects $\angle LDK$ from the definition of $K$, and $\angle MDM' = 90^{\circ}$. This gives us $(K,L;M,M') = -1$. Finally, note that $\angle MPM' = 90^{\circ}$, and therefore $PM$ bisects $\angle LPK$, as desired. Claim 06. $\angle KPC = \angle MAD$ Proof. Note that $\angle KPC = \angle KEC = \angle MEC = \frac{1}{2} \angle MED = \angle MAD$. Finally, to finish this, notice that \[ \angle MPL = \angle KPM = \angle KPC + \angle CPM = \angle DPM + \angle CPM = \angle DPC \]which implies $\angle CPM = \angle DPL$. 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Is this the quickest solution?? Easy angle chase using $AB=BM=CM=CD$ gives $\angle MAD=90 - \angle DMC= \angle MNK=\angle MKN$, thus $AKMN$ is inscribed in a circle. Claim:$(AMD)$ is the Apollonius circle for segment $LK$ and ratio $\frac{LM}{MK}$. Proof:Since we have that $AM$ is an angle bisector in $LAK$ and $DM$ is an angle bisector in $LDK$, because $N$ and $K$ are reflections across $MD$, we get that $(AMD)$ is indeed the Apollonius circle for segment $LK$ and ratio $\frac{LM}{MK}$. $P$ lies on $(AMD)$ so we get that $\angle LPM=\angle KPM$. Now since $CKPN$ is inscribed as well we get the desired result: $$\angle DPL =\angle MPL - \angle MPD = \angle MPL - \angle MAN = \angle MPL - \angle CNK = \angle KPM - \angle CPK = \angle CPM$$

Perhaps a very silly solution but it was the first that occurred to me. Let $O$ the center of $\odot (AMD)$. Claim : $O$ is the reflection of $C$ on $MD$ and $O \in MK$. Proof: It's enough to prove that $\angle DCM = 2\angle DBM$. Let $M'$ a point such that $BDM'M$ is a parallelogram so $MM'CB$ is also a parallelogram $\implies M'C=BM$ then $C$ is the circumcenter of $\odot (MDM') \implies \frac{\angle DCM}{2}= \angle DM'M = \angle DBM$, as desired. $\square$ Let $\Psi$ the inversion with center $O$ and radio equals to radio of $\odot (ADM)$. First: $\angle ODL = \angle DNM$ (Because $OD \parallel CM$) and also $\angle DNM = \angle DKM$ so $\angle DKM = \angle ODN \implies \Psi(K)=L$ By Claim we have that $NK$ and $OC$ have the same perpendicular bisector so $O \in \odot (COK)$ and: $$P = \odot (AMD) \cap \odot (CNK) \implies P=\Psi(P)= \odot (AMD) \cap \Psi(C)L$$By angleS we have that $\Psi(C)$ is the circumcenter of $\odot (MOD)$. The rest is angles chasing, let $\angle MAD= \beta, \angle MOP = \theta, \angle LPO = \alpha$ Then: $$\angle MOD= 2\angle MAD = 2\beta \implies \angle DOP = \theta + 2\beta \implies \angle OPD =90-( \beta + \frac{\theta}{2} )\implies \angle DPL = 90 - \alpha - \beta - \frac{\theta}{2}$$ And also: $\angle LOC = \frac{\angle MOD}{2}= \beta \implies \angle POC = \beta + \theta$ but $PO^2 =O\Psi(C).OC \implies \angle OCP = \angle OPL = \alpha$. Finally: $$\angle OPC = 180 - \angle POC - \angle OCP = 180 - \theta - \beta - \alpha $$$$\angle MPC= \angle OPC - \angle OPM = 180 - \theta - \beta - \alpha - (90 - \frac{\theta}{2}) = 90 - \alpha - \beta - \frac{\theta}{2}$$This implies $\angle MPC = \angle DPL$, as desired. $\blacksquare$ Psdt: 100th Post!

mathematics2004 wrote: In parallelogram $ABCD$ with acute angle $A$ a point $N$ is chosen on the segment $AD$, and a point $M$ on the segment $CN$ so that $AB = BM = CM$. Point $K$ is the reflection of $N$ in line $MD$. The line $MK$ meets the segment $AD$ at point $L$. Let $P$ be the common point of the circumcircles of $AMD$ and $CNK$ such that $A$ and $P$ share the same side of the line $MK$. Prove that $\angle CPM = \angle DPL$. We define and calculate the angle.$$\angle MBC = \angle MCB = \alpha, \angle MDC = \beta \implies MCD = 180 -2\beta,\angle MBA = 2\beta - 2\alpha \implies \angle MAB = 90 + \alpha - \beta,\angle MAD = 90 -\beta \implies \angle MPD = 90 -\beta $$$$\angle MND =\alpha,\angle MDN = \beta - \alpha \implies \angle NMD = 180 - \beta \implies \angle MNK = \angle CNK = \angle CPK = 90 -\beta$$So we have the following expression. $$\boxed{\angle MPD = \angle CPK = 90 -\beta}$$So we need to prove that $PM$ is the bisector of $\angle LPK$.For this it is enough to prove that $\boxed{\frac{PL}{LM} = \frac{PK}{MK} = \frac{PK}{MN}}$. $$\frac{MN}{LM}= \frac{PK}{PL} \iff (\frac{MN}{LM})^2= (\frac{PK}{PL})^2$$$\angle MLN =2\beta - \alpha$.For the $MLN$ triangle. (*) $$\frac{MN}{LM} = \frac{sin(2\beta - \alpha)}{sin \alpha} \implies (\frac{MN}{LM})^2 =(\frac{sin(2\beta - \alpha)}{sin \alpha})^2$$$\angle DAP = \gamma,\angle AMD = 90 + \alpha \implies \angle APD = 90 + \alpha \implies \angle PDA = \angle PDL =90 - \alpha - \gamma $ Let $R$ be the radius of $(AMD)$. $$PL^2 = DP^2 + DL^2 - 2DP\cdot DL \cdot cos(90 - \alpha - \gamma)$$$DP = 2R sin\gamma,DM =2R cos \beta,DN = \frac{Rsin 2\beta}{sin \alpha} $ $$\frac{DL}{sin \beta } = \frac{DM}{sin (2\beta - \alpha)} \implies DL = \frac{Rsin 2\beta}{sin (2\beta - \alpha)}$$$$PL^2 = R^2(4(sin\gamma)^2 + (\frac{sin2\beta}{sin (2\beta - \alpha)})^2 - 2 \cdot 2sin \gamma \cdot \frac{sin2\beta}{sin(2\beta - \alpha)} \cdot sin(\alpha + \gamma)) $$$$\angle PDK = \angle PDM + \angle KDM = \gamma + \beta - 90 + \beta - \alpha = \gamma + 2\beta - \alpha - 90$$$$PK^2 = DK^2 + DP^2 - 2 \cdot DP \cdot DK cos(\gamma + 2\beta - \alpha - 90) = DN^2 + DP^2 - 2 \cdot DP \cdot DN cos(\gamma + 2\beta - \alpha - 90)= R^2((\frac{sin2\beta}{sin\alpha})^2 + 4(sin\gamma)^2 - 2 \cdot 2sin \gamma \cdot \frac{sin 2\beta }{sin \alpha} \cdot cos(\gamma + 2\beta - \alpha - 90)$$$$(\frac{PK}{PL})^2= \frac{(sin2\beta)^2+ 4 \cdot (sin\gamma)^2 \cdot (sin\alpha)^2 - 4 \cdot sin \gamma \cdot sin\alpha \cdot sin 2\beta \cdot cos(\gamma + 2\beta - \alpha - 90)}{(sin2\beta)^2+ 4 \cdot (sin\gamma)^2 \cdot (sin2\beta- \alpha)^2 - 4 \cdot sin \gamma \cdot sin2\beta \cdot sin (2\beta - \alpha) \cdot sin(\alpha+ \gamma)} \cdot (\frac{sin(2\beta - \alpha)}{sin \alpha})^2 =\frac{(sin2\beta)^2+ 4 \cdot (sin\gamma)^2 \cdot (sin\alpha)^2 - 4 \cdot sin \gamma \cdot sin\alpha \cdot sin 2\beta \cdot cos(\gamma + 2\beta - \alpha - 90)}{(sin2\beta)^2+ 4 \cdot (sin\gamma)^2 \cdot (sin2\beta- \alpha)^2 - 4 \cdot sin \gamma \cdot sin2\beta \cdot sin (2\beta - \alpha) \cdot sin(\alpha+ \gamma)} \cdot (\frac{MN}{LM})^2$$hence it suffices to prove the following.$$(sin2\beta)^2+ 4 \cdot (sin\gamma)^2 \cdot (sin\alpha)^2 - 4 \cdot sin \gamma \cdot sin\alpha \cdot sin 2\beta \cdot cos(\gamma + 2\beta - \alpha - 90)= (sin2\beta)^2+ 4 \cdot (sin\gamma)^2 \cdot (sin2\beta- \alpha)^2 - 4 \cdot sin \gamma \cdot sin2\beta \cdot sin (2\beta - \alpha) \cdot sin(\alpha+ \gamma)$$which is also equally strong as follows $$sin\gamma \cdot (sin2\beta- \alpha)^2 - sin2\beta \cdot sin (2\beta - \alpha) \cdot sin(\alpha+ \gamma) = sin\gamma \cdot (sin\alpha)^2 - sin\alpha \cdot sin 2\beta \cdot sin(\gamma + 2\beta - \alpha ) $$.$$sin(\alpha+ \gamma ) = sin \alpha \cdot cos \gamma + sin \gamma \cdot cos \alpha$$.$$sin(\gamma+ 2\beta - \alpha ) = sin \gamma \cdot cos(2\beta -\alpha) + cos\gamma \cdot sin(2\beta - \alpha)$$using these two formulas, let us look at the numbers before $sin \gamma$ above.$$(sin2\beta- \alpha)^2 - sin2\beta \cdot sin (2\beta - \alpha) \cdot cos \alpha = (sin\alpha)^2 - sin \alpha \sin 2\beta cos(2\beta - \alpha)$$.$$(sin2\beta- \alpha)^2 - (sin\alpha)^2 = (sin(2\beta- \alpha) - sin \alpha)( sin(2\beta- \alpha) + sin \alpha) = 2sin (\beta - \alpha)cos \beta \cdot 2sin \beta cos(\beta - \alpha) = sin2\beta \cdot sin(2\beta - 2\alpha) = sin 2\beta \cdot sin(2\beta - \alpha - \alpha) = sin2 \beta \cdot(sin(2\beta - \alpha) \cdot cos\alpha - sin 2\beta sin \alpha cos(2\beta - \alpha) $$. So the numbers before $sin \gamma$ are equal. Now let's look at the numbers before $cos \gamma$ $$-sin2\beta \cdot sin(2\beta - \alpha) \sin \alpha = - sin \alpha \cdot (sin 2\beta) sin(2\beta - \alpha)$$.The proof is over.$\blacksquare$

Let $C'$ be the reflection of $C$ across $MN$. Note that the circumcenter of $CKN$ lies on $MD$, so thus $C'NKC$ is cyclic. Now, we have that since $C'MCD$ and $ABCD$ are both parallelograms, so is $ABMC'$. Thus, we have that $AC' = MC' = DC'$, and so $C'$ is the circumcenter of $AMD$. Now, note that $\angle{DAC'} = \angle{ADC'} = \angle{DNC} = \angle{C'KD}$. Thus, this gives us that $AC'DK$ is cyclic. Hence, by radical axis we find that $L \in PQ$. Now, let $R$ be the midpoint of arc $MD$ on $(AMD)$. Note that since $C"MCD$ is a rhombus, we have that $R \in CC'$; thus, by Fact 5 we have that $R$ is the incenter of $CPQ$. This thus gives us that $PR$ bisects both $\angle{MPD}$ and $\angle{CPQ}$, so thus $\angle{CPM} = \angle{DPL}$, as desired!

Nice one Let $\angle MBA = x$, $\angle CBM = y$, $\angle MNK = z$ and $\angle MDN = a$. Claim : $ANMK$ is cyclic. Proof : $\angle NAM = \angle A - \angle BAM = \angle A - \angle 90 + \frac{x}{2} = \frac{\angle A + y }{2}$. $\angle CDK = \angle CDM - a = \angle CMD - a = y \implies MK || CD$ so $\angle CMK = \angle MCD = \angle A - y = 2\angle NAM$ so $z = \angle NAM \implies ANMK$ is cyclic. Note that by Radical Axis Theorem we have $P,L,Q$ are collinear where $Q$ is second intersection of $AMD$ and $CNK$. Note that $\frac{KA}{AL} = \frac{KM}{ML} = \frac{KD}{DL}$ so $M$ is incenter of $DKA$ so $\angle KPM = \angle MPL$ we need to prove $\angle CPD = \angle MPL$. $\angle CPD = \angle CPQ - \angle DPQ = \angle KPQ - \angle DAQ - z = 2\angle MPL - \angle DAQ - z$ so we must prove $\angle DAQ + z = \angle MPL$ which is true cause $\angle MPL = \angle MPQ = \angle MAQ = \angle DAQ + z$. we're Done.

Nice.

This is cool. Let $F$ be the reflection of $C$ over $MD$.By reflections ,we have $K-M-F$ as $N-M-C$.Also $\angle NFM=\angle MCK$,so $(CNKF)$.Now $FM=CD=AB$,and is also parallel so $FA=BM=FD$,so $F$ is the center of $(AMD)$.Now $\angle DAF= FDL=KDC=DKF$,so $(ADFK)$,hence $L$ lies on the radical axis of the two circles.If their other intersection is $E$,then we have that $CF$,is the angle bisecotr of $PCE$ and $MCD$,so $CM$ and $CD$ are isogonal in $PCE$,or $\angle PCD=\angle NCE=\angle NPL$.We are done by isogonal line lemma in triangle $CPL$ as the point at infinity along $DC$ and point $N$ are isogonal in $CPL$.

Note that $O$ be circumcenter of $(AMD)$ and $Q = (AMD) \cap (CNK)$(not equal $P$) Let $\angle MBC=\angle MCB = a.$ Then, $\angle MKD=\angle MND=\angle MCB=a.$ Let $\angle BAM=\angle BMA= b.$ Then, $\angle MAD=b-a , \angle MCD=2b-2a.$ Claim1: $AKMN$ is cyclic. Proof.$\angle CMD=90+a-b \implies \angle MDA=90-b \implies \angle KND=\angle NKD=b \implies \angle MKN=b-a=\angle MAN$ Claim2: $M$ is incenter of $\triangle AKD.$ Proof. By claim2 we have $\angle AKM=\angle MND =a = \angle DKM.$ So $KM$ and $DM$ are angle bisectors in $\triangle AKD$ Claim3: $K,M,O$ are collinear and $KAOD$ is cyclic. Proof. These are well-known lemmas about incenter. So note that $\angle DAO=\angle ODA=a, \angle AOM=\angle ADK=180-2b, \angle MOD=2b-2a.$ Thus $ABMO$ and $DCMO$ are rhombus. Claim4: $L \in PQ$ Proof. Since $KOAD$ is cyclic by radical lemma we have $L$ lies on radical axis of $(AMD)$ and $(CKN).$ Hence $L$ lies on $PQ.$ Note that by symmetry in $MD$ we have $KCON$ is isosceles trapezoid.Thus $ O \in (KCN).$ Since $MOD=2b-2a \implies MOC=b-a \implies KOC=b-a \implies KPC=b-a.$ Also $\angle MPD=\angle MAD=b-a.$ Let $\angle DPL=c$, so it suffices to show that $\angle CPM=c$ $$\angle CPM=\angle KPQ-(\angle KPC+\angle MPD+\angle DPL)=\angle KOQ-(2b-2a+c)=\angle MOD+\angle DOQ-(2b-2a+c)=\angle DOQ-c=2\angle DPQ-c=c$$ So we are done!

Let $\overline{KM}\cap (CNK)=O\neq K$ and $(CNK)\cap (AMD)=V\neq P$ and . Quick angle and side chasing gives that $CNOD$ and $BMOA$ are rhombi, so $O$ is the center of $(AMD)$. Now note that $KNMA$ is cyclic because $$\angle{MAN}=\angle{MAD}=\frac{1}{2}\angle{MOD}=\angle{NKM}.$$Now, combining this with the similarity of triangles $MNL$ and $OLD$ gives $$\frac{LA}{LK}\overset{(KNMA)}{=} \frac{LM}{LN}\overset{\triangle{MNL}\sim \triangle{OLD}}{=}\frac{LO}{LD}\Rightarrow LA\cdot LD=LO\cdot LK.$$So $OADK$ is cyclic and by $PoP$, $L$ lies on the radical axis of $(CNK)$ and $(ADM)$, i.e. $L\in \overline{PV}$. Now let $R$ be the point diametrically opposite to $M$ with respect to $(ADM)$. Since $\angle{MDR}=90$ and since $\overline{DM}$ bisects the angle $\angle{NDK}$, we have $(K,L;M,R)=-1$. Since $\angle{RPM}=90$ too, our previous cross-ratio gives that $\overline{PM}$ bisects the angle $\angle{LPK}$. Finally, note that $\angle{KPC}=\angle{KOC}=\frac{1}{2}\angle{MOD}=\angle{MPD}$ and thus $$\angle{DPL}=\angle{MPL}-\angle{MPD}=\angle{MPK}-\angle{KPC}=\angle{CPM}$$and we are done.

Note that $KM \parallel CD$ by angle chasing, so if $O$ is the reflection of $C$ over $MD$ the $K, M, L, O$ are collinear. Furthermore, $M$ is the circumcenter of $\triangle{BCO}$, and $(AMD)$ is the circle centered at $O$ passing through $M, D$. So, the problem reduces to as follows: Given rhombus $OMCD$, point $N$ is chosen on line $CM$; if $L = OM \cap ND$ and $P \in (CNO)$ such that $OM = OP$ then $\angle{MPL} = \angle{DPC}$. Let $P'$ be such that $\triangle{NP'L} \sim \triangle{NPD}$; then $\angle{DPC} = \angle{LP'M}$ so it suffices to show that $MLP'P$ is cyclic, which by angle chasing is equivalent to $\angle{OPN} = \angle{MPD}$, which is obvious because both are equal to $\frac{\angle{MCD}}{2}$. $\square$

Cute problem!

Claim 1: $AMNK$ ia cyclic Claim 2: $M$ is the incenter of $ADK$

Then appolonius circle with angle chase finishes.

Angle chase gives $MKL\parallel CD$, and that the circumcenter $E$ of $AMDP$ is on $MKL$. Since $E,A,D,K$ are concyclic, PoP gives $L$ is on radical axis of $(AMD)$ and $(CKN)$, so their radical axis is $LP$. Let the other intersection of the two circles be $L'$, which lies on $PL$. Since $E$ and $C$ are reflections over $DM$, the reflection $Q$ of $P$ over $DM$ lies on $\Gamma = (CKPNEL')$. This is all we need to know about the configuration. Now to finish let $X$ be on $\Gamma$ such that $PX\parallel CD$. We have $EX=PK=NQ$, and $EL'=CD=CQ$, so addition gives $L'X=CN$. This means that the pairs $(PN,P\infty_{CD})$ and $(PC,PL)$ are isogonal, so DDIT from $P$ on trapezoid $CDLM$ gives that $(PC,PL)$ and $(PM,PD)$ are isogonal as well.