In triangle $A B C$, angle $B$ is equal to $120^{\circ}, A B=2 B C$. The perpendicular bisector of side $A B$ intersects $A C$ at point $D$. Find the ratio $A D: D C$.

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Let $M$ be the midpoint of side $AB$.

First method. Drop a perpendicular $CH$ to line $AB$ (see the left diagram). In the right triangle $BHC$, $\angle HBC = 60^\circ$, so

$BH = \frac{1}{2}BC = \frac{1}{4}AB = \frac{1}{2}AM$. Therefore, $HM : MA = 3 : 2$. Since $MD \parallel CH$, by Thales' theorem, $CD : DA = HM : MA = 3 : 2$.
![](https://cdn.mathpix.com/cropped/2024_05_06_7adbd1f2e5e559c8684fg-13.jpg?height=902&width=1874&top_left_y=1938&top_left_x=114)

Second method. Extend segment $DM$ to intersect the extension of side $BC$ at point $K$ (see the right diagram). Since point $K$ lies on the perpendicular bisector of $AB$, $KA = KB$. In the isosceles triangle $AKB$, $\angle ABK = 60^\circ$, so this triangle is equilateral. Therefore, $KC = KB + BC = \frac{1}{2}AB + AB = \frac{3}{2}AB = \frac{3}{2}KA$. The height $KM$ of triangle $AKB$ is also its angle bisector, so $KD$ is the angle bisector of triangle $AKC$. Consequently, $CD : DA = KC : KA = 3 : 2$.

## Answer

$2 : 3$