5. In trapezoid $A B C D$ with bases $A D$ and $B C$, the angle $B C D$ is known to be $120^{\circ}$. A circle with radius 1 is inscribed in this angle, passing through points $A, B$ and $D$. Find the area of triangle $A B D$. Answer: $\frac{\sqrt{3}}{4}$.
Solution: Triangle BCD is isosceles, the angles at the base are $(180-120)/2=30^{\circ}$. Therefore, $\angle B D A=$ $\angle C B D=30^{\circ}$. On the other hand, $\angle B O D=$ $360-90-90-\angle B C D=60^{\circ}$. Then the angle

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$\angle B A D=30^{\circ}$ as an inscribed angle. Sides $A B$ and $BD$ are equal to the radius of the circle, so the area is $\frac{1}{2} A B \times B D \times \sin \angle B=\frac{\sqrt{3}}{4}$.