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 3 of 4: Locate

R

on the circle through

A,M,H,N

Previous step Next step

R=\text{arc midpoint of }MN\text{ not containing }A\text{ on circle }(AMHN) \implies A,M,R,N\text{ concyclic},\ RM=RN

Detailed analysis

Since $OM=ON$ , triangle $OMN$ is isosceles, so the bisector of $angle MON$ is the perpendicular bisector of $MN$ . On the circle $(AMHN)$ , the bisector of the inscribed angle $angle MAN$ passes through the midpoint of the arc $MN$ not containing $A$ . Both defining bisectors therefore meet at that arc midpoint; hence $R$ is it and $RM=RN$ .