Thai Mathematical Olympiad

Observe that $AB\perp NP$. Thus, $M$ is the circumcenter of $△ANP$ and hence $MN=MP$. It can also be seen that $MN\parallel PQ$. From the power of the point $M$, $P{M}^2=M{A}^2=ME\cdot MC$. So, $\frac{PM}{ME}=\frac{MC}{PM}$ and hence $△PME\sim △CMP$. Thus, $M\hat{P}E=M\hat{C}P$. Since $O,A,P,B$ are cyclic, the power of the point $N$ tells us that $CN\cdot NE=AN\cdot NB=ON\cdot NP$. Thus, $C,P,E,O$ are also cyclic and hence $E\hat{P}N=N\hat{C}O$. Since $△PAN\sim △POA$, $\frac{PA}{PN}=\frac{PO}{PA}$. Thus, $PN\cdot PO=P{A}^2=PD\cdot PC$. So, $C,D,N,O$ are cyclic and hence $Q\hat{N}P=P\hat{C}O$. We can now see that $$Q\hat{N}P=P\hat{C}O=P\hat{C}M+M\hat{C}O=M\hat{P}E+E\hat{P}N=M\hat{P}N.$$ Thus, $MP\parallel NQ$. Therefore, $MNQP$ is a rhombus. ■