5. The diagonals of quadrilateral $A B C D$ intersect at point $O$. It is known that $A B=B C=$ $=C D, A O=8$ and $\angle B O C=120^{\circ}$. What is $D O ?$
Answer: $D O=8$.

Solution. Mark a point $E$ on the line $A C$ such that triangle $B O E$ is equilateral (Fig. 2). We will prove the equality of triangles $B A E$ and $B C O$. Indeed, since $A B=B C$, triangle $A B C$ is isosceles, and thus $\angle B A C=\angle B C A$. Additionally, note another pair of equal angles $\angle A E B=\angle B O C=120^{\circ}$. Therefore, triangles $B A E$ and $B C O$ are equal by one side and two angles, hence $A E=C O$ and $A O=A E+E O=C O+B O$.

If we mark a point $F$ on the line $B D$ such that triangle $C O F$ is equilateral, then by similar reasoning we get $D O=B O+C O$. Therefore, it follows that $D O=A O=8$.

![](https://cdn.mathpix.com/cropped/2024_05_06_cbe5f382c3da7be3217bg-2.jpg?height=482&width=860&top_left_y=1537&top_left_x=665)

Fig. 2

Criteria. Only the answer - 0 points. Correct additional construction related to the equilateral triangle - 2 points. The equality $A O=C O+B O$ or its equivalent $D O=B O+C O$ is established - 4 points. Complete solution - 7 points.