Problem 1

Let

ABC

be an acute-angled triangle with

AB\ne AC

. The circle with diameter

BC

intersects the sides

AB

and

AC

M

and

N

respectively. Denote by

O

the midpoint of side

BC

. The bisectors of angles

\angle BAC

and

\angle MON

intersect at

R

. Prove that the circumcircles of triangles

BMR

and

CNR

have a common point lying on side

BC

Step 4 of 4: Force a common point on

BC

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In plain words

A directed-angle identity makes the second intersections with $BC$ coincide.

angle CNR+angle RMB=180^circimplies ext{the circles }(CNR) ext{ \quad\text{and}\quad }(BMR) ext{ meet again on }BC

Detailed analysis

Let $K$ be the second intersection of $(CNR)$ with $BC$ . Write $\angle ACB=a$ , $\angle CBA=b$ , and $\angle ONR=\angle RMO=k$ ; the equality follows because $R$ lies on the perpendicular bisector of $MN$ . From the cyclic quadrilateral $AMRN$ and $AR$ bisecting $\angle A$ , one obtains $\angle RMN=90^\circ-\frac a2-\frac b2$ . On the other hand, angle chasing at $M$ gives $\angle RMN=180^\circ-a-b-k$ , hence $a+b+2k=180^\circ$ . Finally $\angle CNR=a+k$ and $\angle RMB=b+k$ , so $\angle CNR+\angle RMB=180^\circ$ . Therefore $K$ also lies on $(BMR)$ , proving the required common point on $BC$ .