[Ratio of areas of similar triangles] Law of Cosines

In trapezoid $ABCD$, where $BC$ and $AD$ are the bases, diagonal $AC$ is the bisector of angle $BAD$, which is $120^{\circ}$. The radius of the circumcircle of triangle $ABD$ is $\sqrt{3}$. Diagonals $AC$ and $BD$ intersect at point $O$. The areas of triangles $AOD$ and $BOC$ are in the ratio 4:1. Find all sides of trapezoid $ABCD$.
Prove that $A D=2 B C=2 A B$.

## Solution

Let $A B=x$. Since triangle $A B C$ is equilateral ($\angle A C B=\angle C A D=\angle C A B=60^{\circ}$), then $B C=$ $A B=x$ and $A C=x$. Since triangles $A O D$ and $B O C$ are similar with a ratio of $\sqrt{4}=2$, then $A D=2 B C=2 x$.

If $R$ is the radius of the circumcircle of triangle $A B D$, then

$$
B D=2 R \sin \angle B A D=2 \sqrt{3} \cdot \frac{\sqrt{3}}{2}=3
$$

By the cosine rule,

$$
B D^{2}=A B^{2}+A D^{2}-2 A B \cdot A D \cos \angle B A D
$$

or

$$
9=x^{2}+4 x^{2}+2 x^{2}
$$

From this, we find that $x=\frac{3}{\sqrt{7}}$.

By the cosine rule from triangle $A C D$, we find that

$$
C D=3 \sqrt{\frac{3}{7}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_ae1b0d3bbbf9aa5c38e8g-08.jpg?height=294&width=746&top_left_y=-1&top_left_x=682)

## Answer

$$
A B=B C=\frac{3}{\sqrt{7}}, C D=3 \sqrt{\frac{3}{7}}, A D=\frac{6}{\sqrt{7}}
$$