8.5. 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 $A D: D C$.
Answer: $A D: D C=2: 3$.

Solution. First method. Let $M$ be the midpoint of side $A B$. Drop a perpendicular $C H$ to line $A B$ (see Fig. 8.5a). In the right triangle $B H C: \angle H B C=60^{\circ}$, then $\angle B C H=30^{\circ}$, so $B H=\frac{1}{2} B C=\frac{1}{4} A B=\frac{1}{2} A M$. Therefore, $H M: M A=3: 2$. Since $M D \| C H$, by the theorem of proportional segments, $C D: D A=H M: M A=3: 2$.

Second method. Let $M$ be the midpoint of side $A B$. Extend segment $D M$ to intersect the extension of side $B C$ at point $K$ (see Fig. 8.5b). Since point $K$ lies on the perpendicular bisector of $A B$, then $K A=K B$. In the isosceles triangle $A K B \angle A B K=60^{\circ}$, so this triangle is equilateral. Then $K C=K B+B C=\frac{1}{2} A B+A B=\frac{3}{2} A B=\frac{3}{2} K A$. The height $K M$ of triangle $A K B$ is also its angle bisector, so $K D$ is the angle bisector of triangle $A K C$. By the property of the angle bisector of a triangle, $C D: D A=K C: K A=3: 2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f5f37e83183bdb040652g-3.jpg?height=300&width=665&top_left_y=1118&top_left_x=313)

Fig. $8.5 \mathrm{a}$

![](https://cdn.mathpix.com/cropped/2024_05_06_f5f37e83183bdb040652g-3.jpg?height=634&width=569&top_left_y=894&top_left_x=1189)

Fig. $8.5 \sigma$

## Grading Criteria:

+ a complete and well-reasoned solution is provided

$\pm$ a generally correct solution is provided, but minor gaps or inaccuracies are present

干 only the correct answer is provided

- the problem is not solved or is solved incorrectly