In Fig. 14.34, AB

Given,

In a parallelogram ABCD,

∠A = 60^∘

So,

∠B = 180 – 60 = 120 (supplementary angles)

∠ABP = ∠PCB = ^∘

∠DPA = ∠BAP = 30^∘ (AB parallel to DC and AP intersects them)

∠DPA = ∠DAP = 30^∘

AD = DP... (i) (proved)

Similarly,

∠BPC = ∠ABP = 60^∘

In triangle BPC , ∠BPC = ∠PBC = 60^∘

BC = CP = AD... (ii) (proved)

BC = AD

From equation (i) and (ii),

CP = DP

DC = 2AD (proved)