To solve this, we are given:

• $\angle ABC$ is a right angle, meaning it equals $90^\circ$.
• The ray $\overrightarrow{BD}$ bisects $\angle ABC$, meaning it divides the angle into two equal parts.

Solution:

If $\angle ABC = 90^\circ$ and $\overrightarrow{BD}$ bisects it, then each part will be: $\text{m}\angle DBC = \frac{\text{m}\angle ABC}{2} = \frac{90^\circ}{2} = 45^\circ.$

Thus: $\boxed{\text{m}\angle DBC = 45^\circ}.$

Let me know if you'd like further clarification or have other questions! Here are some follow-up questions to deepen understanding:

1. What does it mean geometrically when an angle is bisected?
2. How do we generalize the concept of bisecting an angle for any given angle?
3. Can we calculate the complementary angle to $\angle DBC$ in this case?
4. If the right angle were trisected instead, what would be the measure of each resulting angle?
5. How can we use similar triangles to verify angle bisection?

Tip: When dealing with angle bisections, always divide the total angle into equal parts!