1. **Problem 1:** In the circle ABCDE, CE is a diameter. Given \(\angle BAC = 36^\circ\), find \(\angle BDE\). 2. **Key fact:** The angle subtended by a diameter in a circle is a right angle (90°). Also, angles subtended by the same chord are equal. 3. Since CE is a diameter, \(\angle CBE = 90^\circ\) and \(\angle CDE = 90^\circ\). 4. \(\angle BAC\) and \(\angle BDE\) subtend the same arc BE, so they are supplementary to the angles subtended by the diameter. 5. Using the property of cyclic quadrilaterals, \(\angle BDE = 90^\circ - \angle BAC = 90^\circ - 36^\circ = 54^\circ\). **Answer:** \(\boxed{54^\circ}\) 1. **Problem 2:** In circle ABCD, AC is a diameter. E is the intersection of AC and BD. Given \(\angle BAC = 28^\circ\) and \(\angle BEC = 74^\circ\), find \(\angle ACD\) and \(\angle ACB\). 2. Since AC is a diameter, \(\angle ABD = 90^\circ\) and \(\angle ACD = 90^\circ\) because angles subtended by diameter are right angles. 3. \(\angle BEC = 74^\circ\) is an angle formed inside the circle by chords BE and CE. 4. Using the property of intersecting chords, \(\angle BEC = \frac{1}{2}(\angle BAC + \angle BDC)\). 5. Substitute known values: \(74^\circ = \frac{1}{2}(28^\circ + \angle BDC)\) so \(148^\circ = 28^\circ + \angle BDC\). 6. Solve for \(\angle BDC\): \(\angle BDC = 148^\circ - 28^\circ = 120^\circ\). 7. Since \(\angle ACD\) and \(\angle BDC\) subtend the same arc BC, \(\angle ACD = \angle BDC = 120^\circ\). 8. Finally, \(\angle ACB = 90^\circ\) because AC is diameter. **Answers:** \(\angle ACD = 120^\circ\), \(\angle ACB = 90^\circ\)