3. Given a triangle $ABC$. The angle bisector at vertex $A$ intersects side $BC$ at point $D$. Let $E$, $F$ be the centers of the circumcircles of triangles $ABD$, $ACD$. What can be the measure of angle $BAC$ if the center of the circumcircle of triangle $AEF$ lies on the line $BC$?

(Patrik Bak)
Solution. Let $\alpha$ be the size of the angle $BAC$ under investigation, and $O$ the center of the circumcircle of triangle $AEF$. Since angles $BAD$ and $CAD$ are acute, both points $E$ and $F$ lie in the half-plane $BCA$, and thus for the corresponding central and inscribed angles of the chords $BD$ and $CD$ of the circumcircles of triangles $ABD$ and $ACD$ (Fig. 2) we have

$$
|\angle BED|=2|\angle BAD|=\alpha=2|\angle DAC|=|\angle DFC|.
$$

The isosceles triangles $BED$ and $DFC$ are therefore similar (SAS), so $|\angle EDB| = |\angle FDC|$. Since their bases lie on the same line, it follows that $|\angle EDF| = \alpha$. The line $BC$ is thus the external angle bisector at vertex $D$ of triangle $EDF$. This, as is known, passes through the center of the arc $EDF$ of the circumcircle of triangle $EDF$, i.e., the point that lies on the perpendicular bisector of its chord $EF$. This point is, however, the point $O$, which as the center of the circumcircle of triangle $AEF$ lies on the perpendicular bisector of side $EF$ (and by assumption also on $BC$). Specifically, $|\angle FOE| = |\angle FDE| = \alpha$.

![](https://cdn.mathpix.com/cropped/2024_04_17_ed441ed84b472636ca6fg-4.jpg?height=394&width=1100&top_left_y=2067&top_left_x=478)

Fig. 2

Since $AEDF$ is a kite, $|\angle EAF| = \alpha$. By symmetry, it is clear that the line $EF$ separates points $A$ and $O$ (as we know, point $O$ lies on the arc $EDF$, which is symmetric to the arc $EAF$). According to the theorem on inscribed and central angles, the size of the non-convex angle $EOF$ is twice the size of the convex angle $EAF$. Therefore, $360^\circ - \alpha = 2|\angle EAF| = 2\alpha$, from which it immediately follows that $\alpha = 120^\circ$.

On the other hand, it is easy to verify that at least one such triangle exists (Fig. 3): for example, for $|AB| = |AC|$ and $\alpha = 120^\circ$, points $E$ and $F$ are the midpoints of sides $AB$ and $AC$, triangles $DAE$ and $DAF$ are equilateral, and the center of the circumcircle of triangle $AEF$ indeed lies on side $BC$ (it coincides with the midpoint $D$ of side $BC$).

![](https://cdn.mathpix.com/cropped/2024_04_17_ed441ed84b472636ca6fg-5.jpg?height=343&width=534&top_left_y=505&top_left_x=767)

Fig. 3

Answer. The only possible size of angle $BAC$ is $120^\circ$.