Winter Mathematical Competition

a) Since $△AC{C}_1\sim △AB{B}_1$ and $AN$ and $AM$ are medians in these triangles we have $$\asymp AN{C}_1=\asymp AM{B}_1\Rightarrow \asymp QNB=\asymp PMQ,$$ i.e. the points $M,N,P$ and $Q$ are concyclic.

b) If the points $B,C,P$ and $Q$ are concyclic then $\asymp QCP=\asymp QBP$. But $\asymp AC{C}_1=\asymp AB{B}_1$ and hence $$\asymp QCA=\asymp PBA.(1)$$ On the other hand, since $△AC{C}_1$ and $△AB{B}_1$ are similar we have $$\asymp CAQ=\asymp CAN=\asymp BAM=\asymp BAP.(2)$$ Now (1) and (2) imply that $△ACQ\cong △ABP$, whence $$\frac{AC}{AB}=\frac{AQ}{AP}=\frac{AM}{AN}=\frac{AB}{AC}.$$ Thus $A{B}^2=A{C}^2$ and $AB=AC$.