# Three Collinear and Four Concyclic Points

The problem below is a part of a question posted at the CutTheKnotMath facebook page by Cõ Gẫng Lên:

Points $N, P, Q$ are collinear. For a fourth point $M,$ consider circumcircles, $(A)$ of $\Delta MNP$, $(B)$ of $\Delta MPQ$, $(C)$ of $\Delta NPQ.$

Prove that points $M, A, B, C$ are concyclic.

(The applet below illustrates the problem.)

Solution

Points $N, P, Q$ are collinear. For a fourth point $M,$ consider circumcircles, $(A)$ of $\Delta MNP$, $(B)$ of $\Delta MPQ$, $(C)$ of $\Delta NPQ.$

Prove that points $M, A, B, C$ are concyclic.

### Solution

For definiteness sake, let point $P$ lie between $N$ and $Q$. Angles $NPM$ and $MPQ$ are supplementary: $\angle NPM=180^{\circ}-\angle MPQ.$ $\angle NPM$ is inscribed in circle $(A)$ so that central angle $NAM$ is twice as large. In circle $(B)$, $\angle MBQ=2(180^{\circ}-\angle MPQ)=\angle NAM.$ It follows that two isosceles triangles $AMN$ and $BMQ$ are similar. In addition, one can be obtained from the other by a rotation around $M$ followed, if necessary, by a homothety with center $M$.

Under the two trasformations the perpensicular bisector of $MN$ maps onto the perpendicular bisector of $MQ.$ Let $C'$ be the point of intersection of the two bisectors: $C'N=C'M=C'Q,$ implying $C'=C.$ Also, since $C$ is defined by a rotation around $M$, and $MA$ is mapped onto $MB,$ $\angle AMB+\angle ACB=180^{\circ},$ making the four points $M,A,B,C$ concyclic.

In a little different guise and with a different proof the problem appears elsewhere