Ukrainian Mathematical Competitions

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 $△ACB$ is isosceles and the apex angle is $40^{\circ }$. So,

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

Fig. 47

$$\frac{AO}{AB}=\frac{AB}{AC}\Rightarrow AO\cdot AC=A{B}^2=B{O}^2,$$

What had to be demonstrated.