Given : ABCD is a parallelogram.
To Prove : AE = AD.
Construction : Draw a circle which passes through ABC and intersect CD (or CD produced) at E.
Proof : For fig (i)
∠AED + ∠ABC = 180° [Linear pair] ... (ii)
But ∠ACD = ∠ADC = ∠ABC + ∠ADE
⇒ ∠ABC + ∠ADE = 180° [From (ii)] ... (iii)
From (i) and (iii)
∠AED + ∠ABC = ∠ABC + ∠ADE
⇒ ∠AED = ∠ADE
⇒ ∠AD = ∠AE [Sides opposite to equal angles are equal]
Similarly we can prove for Fig (ii) Proved.