** 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.**