Answer:
See below
Step-by-step explanation:
[tex] m\angle ABC = 2\times m\angle AEC = \boxed {60} \degree [/tex]
(By inscribed angle theorem)
[tex] m\widehat{AC} = m\angle ABC = 60\degree [/tex]
(Central angle and its corresponding arc theorem)
[tex] m\widehat {DC} = 180\degree - (60\degree + 80\degree) [/tex]
[tex] m\widehat {DC} = 40\degree [/tex]
[tex] m\angle DBC = m\widehat {DC} = \boxed {40} \degree [/tex]
(Central angle and its corresponding arc theorem)
[tex] m\angle ACE = \boxed {90} \degree [/tex]
(Angle inscribed in a semicircle)
[tex] m\angle DAE = \frac{1}{2}\times 80\degree = \boxed {40}\degree [/tex]
(By inscribed angle theorem)
[tex] m\angle ADE = \boxed {90}\degree [/tex]
(Angle inscribed in a semicircle)
[tex] m\angle FCB = \boxed {90}\degree [/tex]
(By tangent radius theorem)
[tex] m\angle CED = \frac{1}{2}\times 40\degree = \boxed {20} \degree [/tex]
(By inscribed angle theorem)
