Triangle and Geometric Series

Let $|\overline {AB}|=|\overline {BC}|=a, |\overline {BD}|=\frac ar,|\overline {CD}|=ar=2$

Then, $a^2+\dfrac {a^2}{r^2}=a^2r^2=4$

$\implies r^4-r^2-1=0$

$\implies r^2=\dfrac {\sqrt 5+1}{2}$

$\implies a^2=\dfrac {4}{r^2}=2(\sqrt 5-1)$

So, area of $\triangle {ABC}$ is $\dfrac {a^2}{2}=\sqrt 5-1$

Therefore $A=5,B=1,A\times B=\boxed 5$ .

Since $\angle B$ of $\triangle DBC$ is $90^\circ$ , this means that $DC$ is the diameter of the circle, therefore $DC=2$ . Let $AB= a$ and $BC=b$ . Since $a$ , $b$ , and $2$ are in a geometric progression, we have $2a = b^2 \implies b = \sqrt{2a}$ . By Pythagorean theorem ,

$\begin{aligned} a^2 + b^2 & = 4 \\ a^2 + 2a - 4 & = 0 \\ \implies a & = \frac {-2+\sqrt{4-4(-4)}}2 = \sqrt 5 - 1 & \small \blue{\text{Since }a > 0} \end{aligned}$

The area of $\triangle ABC$ , $[ABC] = \dfrac {b^2}2 = a = \sqrt 5 - 1$ , $\implies A+B = 5 + 1 = \boxed 6$ .