GRAPH In ABC right triangle, $ AC=2+\sqrt3$ and $ BC=3+2\sqrt3$ . Circle goes on point $ C$ and its center is on $ AC$ cathetus, $ C$ Cuts the circle in point $ M$ and it touches $ AB$ hypotenuse at point $ D$ . Find the area of shaded part.

Now my solutions and where I’m stuck at:

I found that $ tg <A=\sqrt3$ which means $ <CAB=60^\circ$ then i used formula $ <CAB=\frac{CD-MD}2$ got to $ {\frown} {CD}=150^\circ$ , $ {\frown} {MD}=30^\circ$ . $ <MOD=30^\circ$ $ <OAD=60^\circ$ , so $ OAD$ is a right triangle.

Then bring in ratios: $ AD=x, AO=2x, OD=\sqrt3$

**Now I’m stuck here** $ AD=2x-x\sqrt3$ , why? I might be missing some rule here, but the next one also confused me. We use another formula $ AD^2=AM*AC (1)$ , I understand why we need to use this, but the calculation confuses me. $ x^2=(2x-x\sqrt3)(2+\sqrt3)$ I don’t understand how we come to this, I’d guess it should be like this instead $ (2x-x\sqrt3)^2=x(2+\sqrt3)$ cause it fits the formula $ (1)$