parmenides51 wrote: On the side $AB$ of the triangle $ABC$ mark the points $M$ and $N$, such that $BM = BC$ and $AN = AC$. Then on the sides $BC$ and $AC$ mark the points$ P$ and $Q$, respectively, such that $BP = BN$ and $AQ = AM$. Prove that the points $C, Q, M, N$ and $P$ lie on the same circle. As, $BC+AC>AB$ then $N$ lies between $MB$ and $M$ lies between $AN.$ $\triangle CBM$ is isoscele and $PBN$ is also isoscele. Again $\angle CBM=\angle PBN$ which proves that $\angle CMB=\angle PNB=\angle BPN$. So, $CMNP$ is cyclic. Similarly, $CNMQ$ is cyclic. So, $C,Q,M,N,P$ lie on a circle.