The Right Kite

Let the center of the circumcircle be $O$ and its radius be $r$ , and $BO$ cuts $AC$ at $P$ .

Since the ratio of areas: $\dfrac{|\triangle ABC|}{|\triangle ADC|} = \dfrac{1}{3} \quad \Rightarrow BP = OP = \dfrac{1}{2}r \quad \Rightarrow DP = \dfrac{3}{2}r$ .

We see that $\cos \angle COP = \dfrac{OP}{OC} = \dfrac {\frac{1}{2}r}{r} = \dfrac{1}{2}\quad \Rightarrow \angle COP = 60^\circ \quad \Rightarrow CP = \dfrac{\sqrt{3}}{2}r$ .

Therefore, $\tan \angle CBP = \dfrac{CP}{BP} = \dfrac{\frac{\sqrt{3}}{2}r}{\frac{1}{2}r} = \sqrt{3} \quad \Rightarrow \angle CBP = 60^\circ \quad \Rightarrow \angle ABC = 2 \times \angle CBP = \boxed{120^\circ}$

Let $\angle ABC=\theta$ , $AB=BC=b$ and $AD=CD=r$ . We know that $\angle ABC+\angle ADC=180^\circ$ because $ABCD$ is cyclic, hence $\angle ADC=180^\circ-\theta$ . Also $[ADC]=3[ABC]$ . Using Law of Sines to compute the area we get: $r^2\sin(180^\circ-\theta)/2=3b^2\sin(\theta)/2$ , which implies $r^2=3b^2$ or $\dfrac{r}{b}=\sqrt{3}$ .

Now, note that $AC=2r\sin\left(\dfrac{180^\circ-\theta}{2}\right)$ and $AC=2b\sin\left(\dfrac{\theta}{2}\right)$ . Then: $r\cos\left(\dfrac{\theta}{2}\right)=b\sin\left(\dfrac{\theta}{2}\right)$ .

Finally, we have $\dfrac{r}{b}=\tan\left(\dfrac{\theta}{2}\right)$ , which is $\sqrt{3}=\tan\left(\dfrac{\theta}{2}\right)$ and $\theta=120^\circ$ .

Let $AC\cap BD ={E}$ , $\Delta BEC ∾ \Delta CED \Rightarrow \dfrac{BE}{CE} = \dfrac{EC}{ED} \Rightarrow \dfrac{BE}{ED} = \left(\dfrac{EC}{ED}\right)^2$

And I have, $\dfrac{BE}{ED}=\dfrac{area(blue)}{area(red)} = \dfrac{1}{3}$ $\Rightarrow \left(\dfrac{EC}{ED}\right)^2 = \dfrac{1}{3} \Rightarrow \dfrac{EC^2+ED^2}{ED^2}= \dfrac{4}{3}\Rightarrow \dfrac{DC^2}{ED^2}= \dfrac{4}{3} \Rightarrow \cos{EDC}=\dfrac{\sqrt{3}}{2} \Rightarrow \angle{ABC} = 120^o$

I will just outline my solution... It's quite easy to show that BD is a diameter. Therefore, angle BAD = angle BCD = 90. Let E be the intersection of the diameter BD and the line AC. Using the ratio 3:1 provided, it follows that BE = BD/4 and DE = 3BD/4. Let F be the center of the circumference. Draw the triangle AFC. Can you see that this triangle and the triangle ABC are congruent? Yes, they are! Therefore, angle ABC = angle AFC. Moreover, AFC = 2ADC. Since ABC + ADC = 180 and ABC = AFC, ABC = 2ADC. Finally, ABC + ABC/2 = 180 and angle ABC = 120.

Since the triangles have same base but 1:3 areas, their heights must be 1:3. The half way point of the combined altitude (the diameter) is the center, But it divides the red altitude in 1:2 (the centroid!) So the red triangle must be equilateral, making the desired angle 180-60 = 120. Because the two triangles together form a cyclic quadrilateral.