61. In quadrilateral $A B C D$ with angles $\angle A=60^{\circ}, \angle B=90^{\circ}, \angle C=120^{\circ}$, the point of intersection of the diagonals $M$ is marked. It turns out that $B M=1$ and $M D=2$. Find the area of $A B C D$.
Answer: 9/2. Solution. $\angle B O D=2 \angle B A D=120^{\circ}$. Let $K$ be the midpoint of $B D$. Then $M K=$ $1 / 2, K D=3 / 2, O K=\sqrt{3} / 2$, so $M K: K O=1: \sqrt{3}$ and therefore $\angle M O K=$ $30^{\circ}$. Hence, $\angle D O C=90^{\circ}$, triangle $A C D$ is an isosceles right triangle (see Fig. (18). By the Law of Sines, $3=B D=A C \sin \angle B A D$, from which $A C=2 \sqrt{3}, S(A C D)=$ $(A C)^{2} / 4=3, S(A B C)=S(A C D) \cdot(B M: M D)=3 / 2, S(A B C D)=9 / 2$.