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 $C^{\prime }$ be the reflection of vertex $C$ over the point $O$ (Fig. 31); then $\angle FC{C}^{\prime }=15^{\circ }$ as well. On the other hand, $\angle ABE={\frac{30}{2}}^{\circ }=15^{\circ }$. In conclusion, we see that the arcs $AE$ and $F{C}^{\prime }$ subtend equal inscribed angles on the circumcircle of triangle $ABC$, hence the corresponding arcs are equal. Since $AB$ and $C{C}^{\prime }$ 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

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Alternative solution.

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