Am I wrong? $$BC=BQ\implies\angle BQC<90^\circ\implies\angle CQP>90^\circ\implies CN=CP>CQ>CN$$

@above Sorry I had a typo in the problem statement!

$\angle{BPC}=\angle{CRA}\implies BPNR \text{ is cyclic}\implies CR\cdot BC=CN\cdot CP\implies BC=CP\implies\frac{CR}{BC}=\frac{CN}{CP}\implies RN||BP\implies A\text{ doesn't exist}$

WolfusA wrote: $\angle{BPC}=\angle{CRA}\implies BPNR \text{ is cyclic}\implies CR\cdot BC=CN\cdot CP\implies BC=CP\implies\frac{CR}{BC}=\frac{CN}{CP}\implies RN||BP\implies A\text{ doesn't exist}$ I believe that $BPMR$ is cyclic

Oh yes, a fault in my diagram.

H.M-Deadline wrote: On the sides of $\triangle{ABC}$ points $P,Q \in{AB}$ ($P$ is between $A$ and $Q$) and $R\in{BC}$ are chosen. The points $M$ and $N$ are defined as the intersection point of $AR$ with the segments $CP$ and $CQ$, respectively. If $BC=BQ$, $CP=AP$, $CR=CN$ and $\angle{BPC}=\angle{CRA}$, prove that $MP+NQ=BR$. $ \angle BCQ=\angle CQB=2x $ $ \angle ACP=\angle PAC=y $ $ \angle CPB=\angle CRA==2y $

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Here is a rather EZ solution: To prove that $MP+NQ=BR$ by adding $RC$ to both sides it is equivalent to prove $CQ+MP=BC$ We start by defining point $T$ on the segment $CQ$ such that $BC=CT$, then it is sufficient to prove that $QT=PM$ , since $\triangle BTC$ and $\triangle NRC$ is isosceles we get that $BT$ is parallel to $NR$ so $$\angle{BTC}=\angle{RNC}=\angle{NRC}=\angle{BPC} \implies BTPC \quad is\quad cyclic $$Also we have that $$\angle{CNA}=180^{\circ}-\angle{NRC}=180^{\circ}-\angle{BPC}=\angle{CPA}\implies PNCA \quad is \quad cyclic$$So we have that $$\angle{TPQ}=\angle{TCB}=\angle{BQC}=\angle{TQP}\quad \implies \quad TQ=TP$$Claim: $ \triangle TPC \cong \triangle MPA$

Proof: Click to reveal hidden text

$\implies TP=PM=TQ$ $Q.E.D$

Let $Q'\in{BC}$ such that $QQ'\|{AR}$. Since we have that $\triangle{RCN}$ is isosceles and $NR\|{QQ'}$ we find that $Q'R=QN$ thus it remains to prove that $BQ'=PM$. We have $\angle{QQ'B}=\angle{ARB}=\angle{APM}=180^{\circ}-\alpha$ where $\alpha=\angle{CRA}$ and $\angle{Q'QB}=\angle{RAB}=\angle{PAM}$. Thus we already have $\triangle{BQQ'}\sim\triangle{MAP}$, so we need to prove that $\triangle{BQQ'}\cong\triangle{MAP}$, which is equivalent to $BQ=AM$. Now we know that $BQ=BC$ and $CP=AP$ so we only need to prove that $$\frac{BC}{PC}=\frac{AM}{AP}$$We prove this by some simple appliance of the Law of Sines! In triangles $\triangle{AMP}$ and $\triangle{BPC}$ we have $$\frac{AM}{AP}=\frac{\sin{(180^{\circ}-\alpha)}}{\sin{\angle{AMP}}}\quad ,\quad \frac{BC}{PC}=\frac{\sin{\alpha}}{\sin{\angle{PBC}}}$$And since $MPBR$ is cyclic from the assumption of $\angle{BPC}=\angle{ARC}$ we find that $$\angle{PBC}=\angle{RMC}=\angle{AMP}$$. So we are done!

Take $V\in CB^\rightarrow$ such that $CV=CQ$. $V$ belongs to the segment $BR$ and $VR=QN$. We are left with proving $BV=PM$.$$\angle{BPC}=\angle{CRA}\implies BPMR \text{ is cyclic}\implies BC\cdot CR=CP\cdot CM=AP\cdot CM$$ Menelaos theorem:$$A\in RM\implies 1=\frac{AB}{AP}\cdot\frac{PM}{CM}\cdot\frac{CR}{BR} =\frac{AB}{BC}\cdot\frac{PM}{BR}$$Hence $$PM=\frac{BR\cdot BQ}{AB}$$$$AR||VQ\implies \frac{BV}{BR}=\frac{BQ}{AB}$$So finally $PM=BV$. QED

Unexpectedly easy for Bulgaria. Using Menelaus theorem on $\triangle CQB$ and points $A,N,R$ we have: $$ 1 =\frac{CN}{QN} \cdot \frac{QA}{BA} \cdot \frac{BR}{CR} \implies QN = \frac{QA}{BA} \cdot BR $$where we used also the fact that $CN=SR$. Note that $\angle BPC = \angle CRA$ implies that quadrilateral $MRBP$ is cyclic. On the other hand from PoP then we have: $$ CR \cdot CB = CM \cdot CP \implies \frac{CM}{CR} = \frac{CB}{CP} = \frac{BQ}{PA}$$where we used the fact that $BC=BQ$ and $CP=PA$. Using Menelaus theorem on $\triangle PCB$ and points $A,M,R$ we have: $$ 1 = \frac{PM}{CM} \cdot \frac{CR}{BR} \cdot \frac{BA}{PA} \implies MP = \frac{PA}{BA} \cdot \frac{CM}{CR} \cdot BR = \frac{BQ}{BA} \cdot BR $$Summing together before established results gives us: $$ QN+MP =\frac{QA}{BA} \cdot BR + \frac{BQ}{BA} \cdot BR = \frac{QA+BQ}{BA} \cdot BR = \frac{BA}{BA} \cdot BR = BR $$as desired.

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Using Menelaus theorem on triangle BCQ and points C, N, R we have: AB/BR=AQ/QN (1) <BPC=<CRA => PMRB cyclic. => Power of Point CMxCP=CRxCB (2) Using Menelaus theorem on triangle BPC and points P, M,C (AB/AP)x(MP/CM)x(CR/BR)=1 following by (1) We have (AB/AP)x(MP/CM)x(CR/BR)=(AB/BR)x(MP/CM)x(CR/AP)=(AQ/QN)x(MP/CM)x(CR/AP)=1 => following by (2) We have (AQ/QN)x(MP/BC)=1 => AQ/QN=BC/MP=AQ/QN=(AQ+BC)/(MP+QN)=(AQ+BQ)/(MP+QN)=AB/(MP+QN)=AB/BR => MP+QN=BR. Proved