Let's draw a circle with the center at the point $C$ and radius $BC = CD$. Points $B$ and $D$ are located on this circle, but $\angle BCD = 120^\circ$, so, arc $AD$, that doesn't contain point $A$, is $240^\circ$. That's why its inscribed angle is $120^\circ$. So, point $A$ is located on the circle too. Then it's easy to calculate angles (Fig. 47).

$\angle ACB = 40^\circ$, $\angle DBA = 40^\circ$ -- is subtended by the central angle $\angle ACD = 80^\circ$, because $\triangle ACB$ is isosceles and the apex angle is $40^\circ$. So,

$\angle CAB = \angle ABC = 70^\circ$. So, $\angle AOB = 70^\circ$ that's why $\triangle AOB$ is also isosceles and similar to $\triangle ACB$, that's why we can write the following equation:

Fig. 47

$$\frac{AO}{AB} = \frac{AB}{AC} \Rightarrow AO \cdot AC = AB^2 = BO^2,$$

What had to be demonstrated.