1. **Problem Statement:** Given circle C with points A, B, D, E, F on the circumference and diameter BE, and angles $m\angle EA = 75^\circ$, $m\angle AD = 68^\circ$, and $m\angle DB = 82^\circ$, find the measures of the listed angles. 2. **Key Properties and Formulas:** - The diameter subtends a right angle to any point on the circle, so any angle subtended by diameter BE is $90^\circ$. - The measure of an inscribed angle is half the measure of its intercepted arc. - Angles subtended by the same chord are equal. - The sum of angles around a point is $360^\circ$. 3. **Step-by-step Solutions:** **Given:** $m\angle EA = 75^\circ$, $m\angle AD = 68^\circ$, $m\angle DB = 82^\circ$. **Note:** Since BE is diameter, $m\angle BFE = 90^\circ$ for any point F on the circle. --- **1. Find $m\angle BF$:** - $\angle DB = 82^\circ$ is given. - Since $\angle BFD$ and $\angle DB$ subtend the same arc BD, $m\angle BF = m\angle DB = 82^\circ$. **2. Find $m\angle FE$:** - $\angle FE$ is part of triangle BFE where $\angle BFE = 90^\circ$ (diameter). - Using triangle sum: $m\angle FE = 180^\circ - m\angle BF - m\angle BFE = 180^\circ - 82^\circ - 90^\circ = 8^\circ$. **3. Find $m\angle ECA$:** - $\angle ECA$ is an inscribed angle subtending arc EA. - Given $m\angle EA = 75^\circ$, so $m\angle ECA = 75^\circ$ (angles subtending same arc are equal). **4. Find $m\angle EBA$:** - $\angle EBA$ subtends arc EA. - By same reasoning, $m\angle EBA = 75^\circ$. **5. Find $m\angle BAF$:** - $\angle BAF$ subtends arc BF. - Since $m\angle BF = 82^\circ$, $m\angle BAF = 82^\circ$. **6. Find $m\angle BEF$:** - $\angle BEF$ subtends arc BF. - So $m\angle BEF = 82^\circ$. **7. Find $m\angle FCE$:** - $\angle FCE$ subtends arc FE. - From step 2, $m\angle FE = 8^\circ$, so $m\angle FCE = 8^\circ$. **8. Find $m\angle EFA$:** - Triangle EFA has angles $m\angle EA = 75^\circ$, $m\angle FE = 8^\circ$. - Sum of angles in triangle: $m\angle EFA = 180^\circ - 75^\circ - 8^\circ = 97^\circ$. **9. Find $m\angle ACD$:** - $\angle ACD$ subtends arc AD. - Given $m\angle AD = 68^\circ$, so $m\angle ACD = 68^\circ$. **10. Find $m\angle EAF$:** - $\angle EAF$ subtends arc EF. - From step 2, $m\angle FE = 8^\circ$, so $m\angle EAF = 8^\circ$. **11. Find $m\angle DCF$:** - $\angle DCF$ subtends arc DF. - Since $m\angle DB = 82^\circ$ and $m\angle AD = 68^\circ$, arc DF = arc DB + arc BF. - Using previous results, $m\angle DCF = m\angle DB + m\angle BF = 68^\circ + 82^\circ = 150^\circ$. **12. Find $m\angle FEA$:** - $\angle FEA$ subtends arc FA. - From previous steps, $m\angle FA = 82^\circ$, so $m\angle FEA = 82^\circ$. **13. Find $m\angle BAE$:** - $\angle BAE$ subtends arc BE. - Since BE is diameter, $m\angle BAE = 90^\circ$. **14. Find $m\angle BEC$:** - $\angle BEC$ is central angle subtending arc BC. - Since C is center, $m\angle BEC = 2 \times m\angle BAC$. - $m\angle BAC = 75^\circ$, so $m\angle BEC = 150^\circ$. **15. Find $m\angle AEC$:** - $\angle AEC$ is central angle subtending arc AC. - $m\angle AEC = 2 \times m\angle ABC$. - $m\angle ABC = 68^\circ$, so $m\angle AEC = 136^\circ$. **Final answers:** 1. $82^\circ$ 2. $8^\circ$ 3. $75^\circ$ 4. $75^\circ$ 5. $82^\circ$ 6. $82^\circ$ 7. $8^\circ$ 8. $97^\circ$ 9. $68^\circ$ 10. $8^\circ$ 11. $150^\circ$ 12. $82^\circ$ 13. $90^\circ$ 14. $150^\circ$ 15. $136^\circ$