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 1 of 4: Identify

M,N

as feet of altitudes

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BM\perp AB \text{ (since } \angle BMC=90^\circ\text{)},\ CN\perp AC \text{ (since }\angle BNC=90^\circ\text{)};\ H:=CM\cap BN=\text{orthocenter}

Detailed analysis

Because $M$ lies on the circle with diameter $BC$ , $angle BMC=90^circ$ , hence $CMperp AB$ . Likewise $BNperp AC$ . Thus $M$ and $N$ are the feet of the altitudes from $C$ and $B$ , and $H=CMcap BN$ is the orthocenter.