The medians $A M$ and $B K$ of triangle $A B C$ intersect the circumcircle of the triangle at points $E$ and $F$ respectively, such that $A E: A M=2: 1, B F: B K=3: 2$. Find the angles of triangle $A B C$.

#
Diagonals $BC$ and $AE$ of quadrilateral $ABEC$ are bisected by their point of intersection $M$, which means this quadrilateral is a parallelogram. Since it is inscribed in a circle, it is a rectangle. Therefore, $\angle BAC = 90^{\circ}$. Let $FK = t, BK = 2t, AK = KC = x$. By the theorem of the products of the segments of intersecting chords, $AK \cdot KC = BK \cdot KF$, or $x^2 = 2t \cdot t = 2t^2$, from which $x = t\sqrt{2}$. From the right triangle $ABK$, we find that

$$
\sin \angle ABK = \frac{AK}{BK} = \frac{x}{2t} = \frac{t\sqrt{2}}{2t} = \frac{\sqrt{2}}{2}
$$

so $\angle ABK = 45^{\circ}$. Then $AB = AK = x$. Therefore,

$$
\tan \angle ABC = \frac{AC}{AB} = \frac{2x}{x} = 2
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b86fbfb53af157d03a0ag-50.jpg?height=1041&width=1037&top_left_y=1126&top_left_x=545)

## Answer

$90^{\circ}, \arctan 2, 90^{\circ} - \arctan 2$