ABCE is a cyclic quadrilateral. AED and BCD are straight lines. AC= CD, angle ABC= 45 degrees and angle ACE = 20 degrees. Work out angle ECD

∠ECD  is 110°

Step-by-step explanation:

For a cyclic quadrilateral,

∠ABC + ∠AEC = 180°

Therefore ∠AEC = 180° - 45° = 135°

∠CAE + ∠ACE + ∠AEC = 180° (Angles in a triangle)

Therefore ∠CAE =  180 - 20 - 135 = 25°

∠CAE = ∠CDA = 25° (Base angles of isosceles triangle ΔCAD)

∠CED + ∠AEC = 180°(Sum of angles on a straight line)

Therefore ∠CED + 135° = 180°

∠CED = 180° - 135° = 45°

∠ECD + ∠CED + ∠CDA = 180°

Therefore ∠ECD  = 180° - (∠CED + ∠CDA)

∠ECD  = 180° - (45° + 25°) = 110°

Therefore, ∠ECD  = 110°.

i think is g(1)=-1.

hope that i can

what? that isnt a question

step-by-step explanation: