$\underline{\text { Eeryxoo } H}$.

In triangle $ABC$, $\angle A=57^\circ$, $\angle B=61^\circ$, $\angle C=62^\circ$. Which of the two segments is longer: the angle bisector of $\angle A$ or the median drawn from vertex $B$?
Let $K$ be the midpoint of the arc $ABC$ of the circumcircle of triangle $ABC$, $O$ be the center of this circle; $N$ be the intersection point of $AL$ and $CK$, and $AH$ be the altitude of triangle $AKC$ (see figure). Since $\angle AAN > AH$. We can reason in different ways.

First method. $AH > KM$, because these are the heights of the smaller and larger angles of triangle $AKC$. Therefore, $KM = MO + OK = MO + OB > MB$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8c2fdb390ca80df56784g-11.jpg?height=1166&width=1155&top_left_y=0&top_left_x=454)

Second method. Since $AB > BC$, then $\angle MBC > 1/2 \angle B > 30^\circ$. Draw the perpendicular $MP$ to side $BC$. Then $AL > AH = 2MP > BM$.

## Answer

$AL > BM$.