Problem 3

Let

ABC

be a triangle. Let

M

and

N

be the points in which the median and the angle bisector, respectively, at

A

meet the side

BC

. Let

Q

and

P

be the points in which the perpendicular at

N

NA

meets

MA

and

BA

, respectively, and

O

the point in which the perpendicular at

P

BA

meets

AN

produced. Prove that

QO

is perpendicular to

BC

Step 1 of 3: Project the arc midpoint K onto the three sides via the Simson line

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In plain words

The midpoint $K$ of arc $BC$ ties the angle bisector $AN$ to the midpoint $M$ of $BC$ , and Simson's theorem links $M$ directly to the projections $R$ and $S$ on $AB$ and $AC$ .

K = AN \cap (ABC), \qquad KB = KC, \quad MB = MC \implies KM \perp BC

Detailed analysis

Let the angle bisector $AN$ meet the circumcircle of $\triangle ABC$ again at $K$ , which is the midpoint of the arc $BC$ not containing $A$ . Since $KB = KC$ and $M$ is the midpoint of $BC$ ( $MB = MC$ ), line $KM$ is the perpendicular bisector of $BC$ , so $M$ is the orthogonal projection of $K$ onto $BC$ . Let $R$ and $S$ be the orthogonal projections of $K$ onto lines $AB$ and $AC$ ; by Simson's theorem, $R, M, S$ are collinear on the Simson line of $K$ with respect to $\triangle ABC$ .