In a rhombus ABCD, O is any interior point such that OA = OC. What is five-ninths of the degree measure of ∠DOB?

Question

Accepted Answer

We are told that:

$$
OA = OC
$$

This means that point $O$ is equidistant from $A$ and $C$. Therefore, **$O$ lies on the perpendicular bisector of segment $AC$**.

But in a rhombus, the perpendicular bisector of diagonal $AC$ is **the other diagonal, $BD$**.

Thus, **point $O$ must lie somewhere on diagonal $BD$**.

This implies that:

* Points $D$, $O$, and $B$ lie on a straight line.
* Hence, $\angle DOB = 180^\circ$

Now we calculate:

$$
\frac{5}{9} 	imes \angle DOB = \frac{5}{9} 	imes 180^\circ = \boxed{100^\circ}
$$

---

### **Final Answer: $\boxed{100^\circ}$**

Answer

Answer

Answer