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 2 of 3: Prove that the Simson line RMS is parallel to PQ

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In plain words

Symmetry across the angle bisector $AN$ forces the projections $R$ and $S$ to be mirror images of each other, so the line $RS$ is automatically perpendicular to $AN$ .

AR = AS,\ KR = KS \implies RS \perp AK \implies RMS \parallel PQ

Detailed analysis

Since $K$ lies on the bisector of $\angle BAC$ and $\angle ARK = \angle ASK = 90^\circ$ , the right triangles $\triangle ARK$ and $\triangle ASK$ are congruent, so $ARKS$ is a kite with $AR = AS$ and $KR = KS$ . Therefore its diagonal $RS$ is perpendicular to its symmetry axis $AK$ (which is line $AN$ ). Because $PQ \perp AN$ by hypothesis, the Simson line $RMS$ is parallel to $PQ$ .