(i). Given: ABCD is a parallelogram.

To Prove:

Proof: In ∆DMC and ∆NMB,

∠DMC = ∠NMB [∵ they are vertically opposite angles]

∠DCM = ∠NBM [∵ they are alternate angles]

∠CDM = ∠MNB [∵ they are alternate angles]

By AAA-similarity, we can say

∆DMC ∼ ∆NMB

So, from similarity of the triangle, we can say

Hence, proved.

(ii). Given: ABCD is a parallelogram.

To Prove:

Proof: As we have already derived

Add 1 on both sides of the equation, we get

⇒

⇒ [∵ ABCD is a parallelogram and a parallelogram’s opposite sides are always equal ⇒ DC = AB]

⇒

Hence, proved.