1. **Problem Statement:** In the circle with center A, BD is a diameter, and $m\angle BAC = 85^\circ$. We need to find: (a) An inscribed angle $\angle BDE$ (b) A semicircle: arc $DE$ (c) A major arc: arc $DCB$ (d) $m\angle BEC$ (e) $m\widehat{BC}$ 2. **Key Concepts:** - A diameter subtends a semicircle (180°). - An inscribed angle is half the measure of its intercepted arc. - The central angle $\angle BAC$ measures 85°. - The arc measure intercepted by $\angle BAC$ is also 85°. 3. **Step (a): Inscribed angle $\angle BDE$** - $\angle BDE$ intercepts arc $BE$. - Since $BD$ is a diameter, $\angle BDE$ is an inscribed angle subtending arc $BE$. - $\angle BAC$ is central and measures 85°, so arc $BC$ measures 85°. - Arc $BE$ includes arc $BC$ plus arc $CE$. - But $\angle BDE$ subtends arc $BE$, and since $BD$ is diameter, $\angle BDE$ is an inscribed angle subtending a semicircle (180°). - Therefore, $m\angle BDE = \frac{1}{2} \times 180^\circ = 90^\circ$. 4. **Step (b): Semicircle $DE$** - Since $BD$ is diameter, arc $DE$ is a semicircle (half the circle). - So arc $DE$ measures 180°. 5. **Step (c): Major arc $DCB$** - The circle is 360°. - Arc $DCB$ is the major arc opposite to minor arc $DB$. - Since $BD$ is diameter, arc $DB$ is 180°. - Therefore, major arc $DCB = 360^\circ - 180^\circ = 180^\circ$. 6. **Step (d): Find $m\angle BEC$** - $\angle BEC$ is an inscribed angle intercepting arc $BC$. - $m\widehat{BC} = 85^\circ$ (from central angle $BAC$). - So $m\angle BEC = \frac{1}{2} \times 85^\circ = 42.5^\circ$. 7. **Step (e): Find $m\widehat{BC}$** - $m\widehat{BC}$ is the measure of the arc intercepted by central angle $BAC$. - Given $m\angle BAC = 85^\circ$, so $m\widehat{BC} = 85^\circ$. **Final answers:** (a) $90^\circ$ (b) arc $DE$ (semicircle) (c) arc $DCB$ (major arc) (d) $42.5^\circ$ (e) $85^\circ$