Let $A B C$ be a right triangle with hypotenuse $B C$ measuring $4 \text{ cm}$. The tangent at $A$ to the circumcircle of $A B C$ meets the line $(B C)$ at point $D$. Suppose that $B A = B D$. Let $S$ be the area of $A C D$, expressed in square centimeters. Calculate $S^{2}$.
Answer: 27. $\alpha=\widehat{A D B}=\widehat{B A D}=\widehat{A C B}=30^{\circ}$ because $\widehat{A B C}=2 \alpha$. Therefore, $A B=B D=$ radius of the circle $=2, C D=6$ and the height is $\sqrt{3}$.