In Fig. 14.34, ABCD is a parallelogram in which ∠A=60°. If the bisectors of ∠A and ∠B meet at P, prove that AD=DP, PC=BC and DC=2AD.
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)