# Stuck on step of the solution Circle theorems/Triangles

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 $$tg which means $$ then i used formula $$ got to $${\frown} {CD}=150^\circ$$, $${\frown} {MD}=30^\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 $$(1)$$