Problem 8.3

In triangle $A B C$, angle $\mathrm{B}$ 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 $C D: D A$.

## Number of points 7

#
# Answer:

$3: 2$

## Solution

We need the ratio of segments lying on the same straight line. We apply Thales' theorem. The angle $CAB$, one straight line $DM$ is already there (and it passes through the point $D$ we need, which is another confirmation that it is appropriate to adapt to Thales' theorem here). We need another straight line, parallel to $DM$, and it is clear that it should pass through point $C$. We draw the line $CH$, parallel to $DM$. We get a right triangle $CBH$ with an angle of $60^{\circ}$. Then its second angle is $30^{\circ}$. The leg opposite the $30^{\circ}$ angle is half the hypotenuse. Therefore, $HM: MA = 3: 2$. Then, by Thales' theorem, the same ratio applies to the segments $CD$ and $DA$.

![](https://cdn.mathpix.com/cropped/2024_05_06_b2e5414d7470d94f852cg-2.jpg?height=289&width=511&top_left_y=1072&top_left_x=1412)