Estonian Mathematical Olympiad

Let $\angle ABC = 30^\circ$ and $\angle BCA = 90^\circ$. Let $O$ be the circumcenter of triangle $ABC$; it is also the midpoint of the hypotenuse $AB$. We have $\angle OAC = \angle BAC = 60^\circ$, and since $AO = CO$, it follows that $\angle OCA = 60^\circ$ (Fig. 30). Therefore, $\angle FCO = 60^\circ - \frac{90^\circ}{2} = 15^\circ$. Let $C'$ be the reflection of vertex $C$ over the point $O$ (Fig. 31); then $\angle FCC' = 15^\circ$ as well. On the other hand, $\angle ABE = \frac{30}{2}^\circ = 15^\circ$. In conclusion, we see that the arcs $AE$ and $FC'$ subtend equal inscribed angles on the circumcircle of triangle $ABC$, hence the corresponding arcs are equal. Since $AB$ and $CC'$ are diameters, the remaining arcs $BE$ and $CF$ are also equal. Therefore, the corresponding chords $BE = CF$ are equal. Thus, the equality $BE = CF$ can hold even in a non-isosceles triangle.

Fig. 30

Fig. 31

Let $\angle CAB = 60^\circ$ and $AC < AB$, and let $O$ be the circumcenter of triangle $ABC$ (Fig. 32). Then $\angle BOC = 2 \cdot 60^\circ = 120^\circ$. Thus, $\angle COE + \angle EOA + \angle AOF + \angle FOB = 360^\circ - 120^\circ = 240^\circ$. Since $\angle COE = \angle EOA$ and $\angle AOF = \angle FOB$, we have $\angle EOA + \angle AOF = \frac{240^\circ}{2} = 120^\circ$. Consequently, $\angle EOF = \angle BOC$, from which it follows that
$$\angle COF = \angle COE + \angle EOF = \angle BOC + \angle COE = \angle BOE.$$
In conclusion, we see that the central angles subtended by the shorter arcs $BE$ and $CF$ of the circumcircle of triangle $ABC$ are equal, hence the corresponding arcs are equal. Therefore, the corresponding chords $BE$ and $CF$ are equal. Thus, the equality $BE = CF$ can hold in a non-isosceles triangle.

Fig. 32