China Southeastern Mathematical Olympiad

We show that the above conclusion is true for any triangle.

If $△ABC$ is right-angled. The conclusion is obvious. In fact, see Fig. 4. 1, where $\angle ABC=90^{\circ }$. So, the circumcentre $O$ is the midpoint of $AC$, $OA=OB$ and $N=O$. Since $F$ is the midpoint of $CM$, we see that the median line $OF\parallel AM$. Hence $\angle EOF=\angle OBA=\angle OAB=\angle A$.

If $△ABC$ is not right-angled, see Fig. 4. 2 and Fig. 4. 3. Fig. 4. 1 Fig. 4. 2 Fig. 4. 3

First we give a lemma.

Lemma. Let $A$ and $B$ be two points on the diameter $KL$ of circle $\odot O$ with radius $R$, and $OA=OB=a$. See figure. Let $CD$ and $EF$ be two chords passing $A$ and $B$, respectively. Suppose $CE$ and $DF$ intersect $KL$ at $M$ and $N$, respectively. Then $MA=NB$.

Proof of the lemma. As shown in Fig. 4. 4. Suppose that $CD\cap EF=P$. Think of that lines $CE$ and $DF$ intersect $△PAB$. By Menelaus' Theorem, we have Fig. 4. 4 $$\frac{AC}{CP}\cdot \frac{PE}{EB}\cdot \frac{BM}{MA}=1,$$ $$\frac{BF}{FP}\cdot \frac{PD}{DA}\cdot \frac{AN}{NB}=1.$$ Then $$\frac{MA}{NB}=\frac{AC}{BE}\cdot \frac{AD}{BF}\cdot \frac{PE}{PC}\cdot \frac{PF}{PD}\cdot \frac{BM}{AN}.\stackrel{◯}{1}$$ By the Intersecting Chord Theorem, we get $$PC\cdot PD=PE\cdot PF.\stackrel{◯}{2}$$ So $$AC\cdot AD=AK\cdot AL=R^2-a^2=BK\cdot BL=BE\cdot BF.\stackrel{◯}{3}$$ By ①, ③, we have $\frac{MA}{NB}=\frac{MB}{NA}$, that is $$\frac{MA}{NB}=\frac{MA+AB}{NB+AB}=\frac{AB}{AN}=1.$$ Thus, $MA=NB$.