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The problem involves a circle with diameters \( AC \) and \( BD \) that intersect at point \( E \), and these diameters are perpendicular. This setup forms right angles at \( E \) and divides the circle into four right-angled triangles.

Let's go through each question:

### 31. \( m\angle AED \)
Since \( AC \) and \( BD \) are diameters and are perpendicular, they form a right angle at \( E \). Thus:
\[
m\angle AED = 90^\circ
\]

### 32. \( m\overset{\frown}{AB} \)
The arc \( AB \) is intercepted by the central angle \( \angle AEB \), which measures \( 90^\circ \) (since \( \angle AEB \) is part of the perpendicular diameters). Therefore:
\[
m\overset{\frown}{AB} = 90^\circ
\]

### 33. \( m\angle ADC \)
Since \( A \), \( D \), and \( C \) are points on the circle and \( AC \) is a diameter, \( \angle ADC \) is an inscribed angle that intercepts a semicircle (180° arc). By the inscribed angle theorem, an angle that intercepts a semicircle is a right angle:
\[
m\angle ADC = 90^\circ
\]

### 34. \( m\overset{\frown}{BCD} \)
The arc \( BCD \) is the major arc that covers three of the four 90° quadrants in the circle. Thus:
\[
m\overset{\frown}{BCD} = 270^\circ
\]

### 35. \( m\angle BDC \)
This is an inscribed angle that intercepts the arc \( BC \), which is 90°. By the inscribed angle theorem, \( m\angle BDC \) is half of the intercepted arc:
\[
m\angle BDC = \frac{90^\circ}{2} = 45^\circ
\]

### Summary of Answers
- **31.** \( m\angle AED = 90^\circ \)
- **32.** \( m\overset{\frown}{AB} = 90^\circ \)
- **33.** \( m\angle ADC = 90^\circ \)
- **34.** \( m\overset{\frown}{BCD} = 270^\circ \)
- **35.** \( m\angle BDC = 45^\circ \)

Would you like any additional details or explanations?

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**Related Questions:**
1. Why is an inscribed angle half of the intercepted arc's angle?
2. What are the properties of diameters that intersect at right angles in a circle?
3. How does the central angle relate to its intercepted arc in a circle?
4. Can you apply the same concepts if the diameters were not perpendicular?
5. What happens to inscribed angles in a circle when they intercept the same arc?

**Tip:** In a circle, an inscribed angle is always half the measure of the arc it intercepts. This is helpful for quickly finding angle measures based on arc lengths.

Given that AC and BD are diameters of circle O centered at E and are perpendicular, find the measures of the indicated angles and arcs.

Explore the geometry of a circle with perpendicular diameters AC and BD that intersect at the center E. This problem covers the calculation of angle measures such as m∠AED and m∠ADC, as well as arc measures like mAB and mBCD, using inscribed angle theorems and properties of circles. Suitable for high school students.

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