6. Tetrahedron $ABCD$ with acute-angled faces is inscribed in a sphere with center $O$. A line passing through point $O$ perpendicular to the plane $ABC$ intersects the sphere at point $E$ such that $D$ and $E$ lie on opposite sides relative to the plane $ABC$. The line $DE$ intersects the plane $ABC$ at point $F$, which lies inside triangle $ABC$. It turns out that $\angle ADE = \angle BDE$, $AF \neq BF$, and $\angle AFB = 80^\circ$. Find the measure of $\angle ACB$.
Answer: $40^{\circ}$.

Note that point $E$ is equidistant from points $A, B, C$, so its projection onto the plane $A B C$ coincides with the projection of point $O$ onto this plane and is the center of the circumscribed circle of triangle $A B C$.

Consider triangles $A D E$ and $B D E$. They have a pair of equal sides $A E$ and $B E$, a common side $D E$, and equal angles $A D E$ and $B D E$. From the Law of Sines, it follows that these triangles are either equal or the angles $D A E$ and $D B E$ are supplementary to $180^{\circ}$. The first situation is impossible, as in the case of equality of triangles $A D E$ and $B D E$, points $A$ and $B$ are equidistant from any point on side $D E$, but by the condition $A F \neq B F$. Therefore, $\angle D A E + \angle D B E = 180^{\circ}$.

Consider the point $X$ of intersection of the ray $A F$ with the sphere $\Omega$ circumscribed around the tetrahedron $A B C D$. Note that the ray $A F$ lies in the planes $A B C$ and $A E D$, so point $X$ lies on the circumscribed circles of triangles $A B C$ and $A E D$. Point $E$ is equidistant from all points of the circumscribed circle of triangle $A B C$; in particular, $A E = X E$. From the inscribed quadrilateral $A E X D$, it follows that $\angle D A E + \angle D X E = 180^{\circ}$. Since $A E = X E$, $E$ is the midpoint of the arc $A X$ of the circumscribed circle of triangle $A D E$, and thus $\angle A D E = \angle X D E$.

Using the previously derived angle equalities, we conclude that triangles $D B E$ and $D X E$ are equal by the second criterion: $\angle D B E = 180^{\circ} - \angle D A E = \angle D X E, \angle X D E = \angle A D E = \angle B D E$, side $D E$ is common. Since triangles $D B E$ and $D X E$ are equal, vertices $B$ and $X$ are equidistant from any point on side $D E$; in particular, $B F = F X$.

It remains to calculate the angles in the plane $A B C$. Sequentially using the inscribed quadrilateral $A B X C$, the isosceles triangle $B F X$, and the exterior angle theorem for triangle $B F X$, we write

$$
\angle A C B = \angle A X B = \frac{1}{2} \cdot (\angle F X B + \angle F B X) = \frac{1}{2} \cdot \angle A F B = 40^{\circ}
$$

Another solution. Let the ray $A F$ intersect the sphere $\Omega$ circumscribed around the tetrahedron $A B C D$ at point $X$. By construction, the point $E$ satisfies the relation $E X = E A$, which implies that $\angle A D E = \angle E A F$. Similarly, we obtain that $\angle B D E = \angle E B F$, and thus $\angle E A F = \angle E B F$.

Denote the point of intersection of the line $O E$ with the plane $A B C$, which is the center of the circumscribed circle of triangle $A B C$, as $O_{1}$. Then $\angle O_{1} A E = \angle O_{1} B E$.

Consider the trihedral angles $A O_{1} E F$ and $B O_{1} E F$. In them, the planar angles $E A F$ and $E B F$, the planar angles $O_{1} A E$ and $O_{1} B E$, and the dihedral angles at the edges $A O_{1}$ and $B O_{1}$ are right. Therefore, the corresponding trihedral angles are equal. And thus the planar angles $\angle F A O_{1} = \angle F B O_{1}$ are equal. Note that this equality can also be derived from the Law of Cosines for trihedral angles.

The specified equality is possible in two cases: either point $F$ lies on the perpendicular bisector of $A B$ (points $A$ and $B$ are symmetric relative to $F O_{1}$), or point $F$ lies on the circumscribed circle of triangle $A B O_{1}$. The first case is forbidden by the condition $A F \neq B F$, so the second case must hold. Then $\angle A O B = \angle A F B = 80^{\circ}$ and is the central angle for the angle $A C B$ in the circumscribed circle of triangle $A C B$. As a result, we conclude that $\angle A C B = 40^{\circ}$.

## Criteria

The following points are summed.