In `Delta ABC, AB = AC, O `is a point on BC such that BO = CO and OD is perpendicular to AB and OE is perpendicular to AC. If `/_BOD = 30^@` then measure of `/_AOE` is

Options

A. `45^@`
B. `60^@`
C. `75^@`
D. `30^@`

Answer

To solve the problem, we need to analyze the given information about triangle ABC and the points O, D, and E.

### Step-by-Step Solution:

1. **Identify the Triangle and Points**:
- We have triangle ABC where AB = AC (isosceles triangle).
- Point O is on line segment BC such that BO = CO, meaning O is the midpoint of BC.
- OD is perpendicular to AB, and OE is perpendicular to AC.

2. **Understanding Angles**:
- Given that angle BOD = 30 degrees, we need to find angle AOE.

3. **Using Properties of Isosceles Triangle**:
- Since triangle ABC is isosceles with AB = AC, angles ABC and ACB are equal. Let's denote these angles as x.
- The vertex angle A (angle BAC) can be expressed as:
\[ \angle A = 180^\circ - 2x \]

4. **Analyzing Triangle BOD**:
- In triangle BOD, we know:
- Angle BOD = 30 degrees
- Angle OBD = 90 degrees (since OD is perpendicular to AB)
- Therefore, angle ODB can be calculated as:
\[ \angle ODB = 180^\circ - \angle BOD - \angle OBD = 180^\circ - 30^\circ - 90^\circ = 60^\circ \]

5. **Finding Angle AOE**:
- Since OE is perpendicular to AC, angle OEA is also 90 degrees.
- In triangle AOE, we can find angle AOE using the fact that the angles in a triangle sum up to 180 degrees:
\[ \angle AOE + \angle OAE + \angle OEA = 180^\circ \]
- We know that angle OEA = 90 degrees and angle OAE = angle ODB = 60 degrees (from step 4).
- Thus, we have:
\[ \angle AOE + 60^\circ + 90^\circ = 180^\circ \]
- Simplifying this gives:
\[ \angle AOE + 150^\circ = 180^\circ \]
- Therefore:
\[ \angle AOE = 180^\circ - 150^\circ = 30^\circ \]

### Final Answer:
The measure of angle AOE is **30 degrees**.