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 3 of 3: Apply a homothety centered at A to map OQ to KM

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In plain words

The configuration $(O, P, Q)$ is simply a scaled copy of $(K, R, M)$ from center $A$ ; sliding $N$ along the bisector just scales $\triangle OPQ$ without changing the direction of $OQ$ .

\mathcal{H}_A: O \mapsto K \implies P \mapsto R \implies Q \mapsto M \implies OQ \parallel KM \perp BC

Detailed analysis

Consider the homothety $\mathcal{H}_A$ centered at $A$ that takes $O$ to $K$ along line $AN$ . Since $OP \perp AB$ and $KR \perp AB$ , lines $OP$ and $KR$ are parallel, so $\mathcal{H}_A$ maps line $OP$ to line $KR$ and hence takes $P = OP \cap AB$ to $R = KR \cap AB$ . Next, since $PQ \parallel RS$ and $M \in RS$ , $\mathcal{H}_A$ maps line $PQ$ to line $RS$ , taking $Q = PQ \cap AM$ to $M = RS \cap AM$ . Because $\mathcal{H}_A(O) = K$ and $\mathcal{H}_A(Q) = M$ , line $OQ$ is parallel to line $KM$ . Since $KM \perp BC$ , it follows that $QO \perp BC$ .