The measure of the angle between the internal and external bisectors of an angle is

Options

A. `60^@`
B. `70^@`
C. `80^@`
D. `90^@`

Answer

To find the measure of the angle between the internal and external bisectors of an angle, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Angle Bisectors**:
- Let the angle be represented as \( \angle BOD \) where \( O \) is the vertex of the angle and \( B \) and \( D \) are points on the two rays forming the angle.

2. **Define the Angle**:
- Let \( \angle BOD = \theta \).

3. **Internal Bisector**:
- The internal bisector of the angle \( \angle BOD \) divides it into two equal parts. Therefore, the angle formed by the internal bisector and the rays is:
\[ \angle DOH = \frac{\theta}{2} \]

4. **External Bisector**:
- The external bisector of the angle \( \angle BOD \) is the line that divides the exterior angle \( \angle AOB \) (where \( A \) is a point on the extension of ray \( OB \)) into two equal parts. The angle \( \angle AOB \) is:
\[ \angle AOB = 180^\circ - \theta \]
- Thus, the angle formed by the external bisector and the rays is:
\[ \angle EOD = \frac{1}{2}(180^\circ - \theta) = 90^\circ - \frac{\theta}{2} \]

5. **Finding the Angle Between Internal and External Bisectors**:
- The angle between the internal bisector (line \( OH \)) and the external bisector (line \( OE \)) is given by:
\[ \angle EOH = \angle EOD + \angle DOH \]
- Substituting the values:
\[ \angle EOH = \left(90^\circ - \frac{\theta}{2}\right) + \left(\frac{\theta}{2}\right) \]
- Simplifying this gives:
\[ \angle EOH = 90^\circ \]

6. **Conclusion**:
- Therefore, the measure of the angle between the internal and external bisectors of an angle is \( 90^\circ \).

### Final Answer:
The measure of the angle between the internal and external bisectors of an angle is \( 90^\circ \).

---