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

If $\triangle ABC$ 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 $\triangle ABC$ 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 $\triangle PAB$. 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
$$\frac{MA}{NB} = \frac{AC}{BE} \cdot \frac{AD}{BF} \cdot \frac{PE}{PC} \cdot \frac{PF}{PD} \cdot \frac{BM}{AN}. \qquad \textcircled{1}$$
By the Intersecting Chord Theorem, we get
$$PC \cdot PD = PE \cdot PF. \qquad \textcircled{2}$$
So
$$AC \cdot AD = AK \cdot AL = R^2 - a^2 = BK \cdot BL = BE \cdot BF. \qquad \textcircled{3}$$
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, $MA = NB$.