Step One
Find <ACB
Since AB = BC the triangle ABC is isosceles. That means that the angles opposite the equal sides are equal.
Therefore <ACB = <CAB = <40o
Step Two
find angle ABC
ABC + BAC + BCA = 180o All Triangles have 180o
<ABC + 40o + 40o = 180o
<ABC + 80o = 180o
<ABC = 180o - 80o
<ABC = 100o
Step three
Find <ADC
In a cyclic Quadrilateral The angles opposite each other are Supplementary. So because <ABC and <ADC are oppose each other they will equal 180o.
<ABC + <ADC = 180o
100o + <ADC = 180o
<ADC = 180o - 100o
<ADC = 80o
Step Four
Find <ADE
<ADC +< ADE = 180o
<ADE + 80o = 180o
<ADE = 100o