Q. 73.6( 24 Votes )

In Fig. 14.34, AB

Answer :

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)


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