Let $A=\angle{BAC},\theta=\angle{ANM},\phi=\angle{AMN}$. $\angle{QMB}=\angle{BMP}=\angle{PMN}+\angle{NMQ}=\angle{PMN}+2\angle{AMN}=A+2\theta$ $QM:MB=PM:MB=AC:AB$ $\angle{CNQ}=\pi+\angle{ANM}-\angle{MNQ}=\pi+\angle{ANM}-\angle{MNP}=\\ \pi+\theta-\phi=(A+\theta+\phi)+\theta-\phi=A+2\theta.$ $CN:NQ=CN:NP=AC:AB$. Thus the result follows.

Bigwood, looking at my diagram, I can't understand your solution. Could you explain a bit more? Here is my solution: Since $AMPN$ is a parallelogram, $\angle MAN=\angle MPN$ but $\angle MPN=\angle MQN$ hence $\angle MAN=\angle MQN$ and so $MQAN$ is cyclic. Therefore $\angle QMB=180^{\circ}-\angle QMA=180^{\circ}-\angle QNA=\angle QNC$. Therefore it suffices to prove that $\frac{MQ}{MB}=\frac{NC}{NQ}$ in order to prove that $\triangle QMB\sim\triangle CNQ$. Now $MN$ clearly bisects both $PQ$ and $PA$. Therefore $MN||QA$. Hence $QANM$ is a cyclic trapezium, and so is an isosceles cyclic trapezium, meaning $MQ=AN$ and $NQ=AM$. Therefore $\frac{MQ}{MB}=\frac{NC}{NQ}\iff \frac{AN}{MB}=\frac{NC}{AM}\iff \frac{AN}{NC}=\frac{MB}{AM}$. Now since $PM||AC$ we have $\frac{MB}{AM}=\frac{PB}{CP}$. Also since $PN||AB$ we have $\frac{AN}{NC}=\frac{PB}{CP}$, hence $\frac{AN}{NC}=\frac{MB}{AM}$.

Since $AMPN$ is a parallelogram, we have $\measuredangle MQN = -\measuredangle MPN = \measuredangle MAN$, implying that $A, M, N, Q$ are concyclic. Hence, $\measuredangle QMB = \measuredangle QMA = \measuredangle QNA = \measuredangle QNC \; (\star).$ Meanwhile, note that $\triangle MBP \sim \triangle NPC \sim \triangle ABC.$ Hence, \[\frac{QM}{MB} = \frac{PM}{MB} = \frac{CN}{NP} = \frac{CN}{NQ}. \; (\dagger)\]Combining $(\star)$ and $(\dagger)$, it follows from side-angle-side similarity that $\triangle QMB \sim \triangle CNQ$, as desired. $\square$

İs my solution wrong or Why such hard solutions for that easy problem.? Writing Thales two times $\frac{BM}{AB}=\frac{MP}{AC}$ similarly $\frac{CN}{AC}=\frac{NP}{AB}$ and multipliying them yields that $BM\cdot CN=MP\cdot NP$ and replacing $MP,NP$ with $MQ,NQ$ we are done...

I think the hard part is to guess that the triangles are not similarly oriented.I spent a lot time thinking to prove $\triangle{QMN} \sim \triangle{QBC}$ which is wrong. .It is an easy angle chase to show $\angle{QMB}=\angle{QNC}$ and writing Thales as above will complete the solution.

$\triangle BMP$ and $\triangle PNC$ are homothetic and $MN$ goes through center of this homothety. Reflect $P \mapsto Q, C \mapsto D$ about $MN$ to obtain also homothetic triangles $BMQ$ and $PND$. And finally, notice that $\triangle PND = \triangle QNC$ by symmetry.

$\angle MAN=\angle MPN=\angle MQN \Longrightarrow AQNM$ is cyclic. So $\angle BMQ=\pi-\angle AMQ=\pi-\angle ANQ=\angle QNC$ $QN=PN$ and $MQ=MP$, So, since $\triangle MBP \sim NPC$, $\frac{MB}{MQ}=\frac{MB}{MP}=\frac{NP}{NC}=\frac{NQ}{NC}$ , so $\triangle QMB \sim CNQ$.$\blacksquare$

$QAMN$ cyclic

Notice $$\measuredangle MAN=\measuredangle NPM=\measuredangle MQN$$so $AMQN$ is cyclic. Since $\triangle BMP\sim\triangle PNC,$ we see $$QM/MB=MP/MB=NC/NP=NC/NQ.$$Noting $-\measuredangle AMQ=-\measuredangle ANQ$ finishes. $\square$

Clearly $AMPN$ is a paralelogram so its easy to see that $ANMQ$ is a ISLsuscelesTrapezoid so $\angle QMB=\angle CQN$, now by thales spam we have that $$\frac{BM}{MP}=\frac{BA}{AC}=\frac{PN}{NC} \implies \frac{BM}{MQ}=\frac{QN}{NC}$$So from this and the angle we get $\triangle BMQ \sim \triangle QNC$ thus we are done

Rip this took me 40 minutes because I got trolled by the similarity not being in the orientation you would expect and kept trying to prove that AQBC was cyclic, although after I realized that mistake i got it pretty quickly Claim: $\angle QMB=\angle CNQ.$ Note that since $ANPM$ is a parallelogram, $ANMQ$ is a cyclic isoceles trapezoid, so $\angle QMA=\angle QNA$. Therefore, we have $$\angle QMB=\angle CNQ=180-\angle QMA=180-\angle QNA.$$ Let $a=BC,b=AC,c=AB.$ We will now start using barycentric coordinates. Let $$P=(0,p,1-p).$$Then, we have $$M=(1-p,p,0),N=(p,0,1-p).$$This means that $$BM=(1-p)c,CN=pb.$$Also, note that $$QM=AN=(1-p)b,$$and $$QN=AM=pc.$$Then, we have $$BM\cdot CN=QM\cdot QN=p(1-p)bc,$$so by SAS, we have $\triangle BMQ\sim \triangle QNC$ as desired.