
3. Let $A B C$ be a triangle. The $A$-angle bisector intersects $B C$ at $D$. Let $E, F$ be the circumcenters of triangles $A B D, A C D$, respectively. Given that the circumcenter of triangle $A E F$ lies on $B C$, find all possible values of $\angle B A C$. (Patrik Bak)


Solution. Let $O$ be the circumcenter of triangle $A E F$ and denote $\alpha=\angle B A C$. Since $\angle B A D$ and $\angle C A D$ are acute (Fig. 2), points $E, F$ lie in the half-plane $B C A$ and the Inscribed angle theorem yields

$$
\angle B E D=2 \cdot \angle B A D=\alpha=2 \cdot \angle D A C=\angle D F C \text {. }
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_c0cc9d17594ecfc50dfag-14.jpg?height=400&width=1100&top_left_y=1959&top_left_x=518)

Fig. 2

The isosceles triangles $B E D$ and $D F C$ are thus similar and we easily compute that $\angle E D F=\alpha$ and that $B C$ is the external $D$-angle bisector in triangle $D E F$.

Point $O$ lies on $B C$ and on the perpendicular bisector of $E F$. Framed with respect to triangle $D E F$, it lies on the external $D$-angle bisector and on the perpendicular bisector of the opposite side $E F$. Thus it is the midpoint of $\operatorname{arc} E D F$ and $\angle E O F=$ $\angle E D F=\alpha$.

Quadrilateral $A E D F$ is a kite, hence $\angle E A F=\alpha$. Moreover, line $E F$ separates points $A$ and $O$, thus the Inscribed angle theorem implies that the size of the reflex angle $E O F$ is twice the size of the convex angle $E A F$. This yields $360^{\circ}-\alpha=2 \cdot \alpha$ and $\alpha=120^{\circ}$.

Answer. The only possible value is $\angle B A C=120^{\circ}$.
