Question

In the given diagram ABCD is a parallelogram. APD and BQC are equilateral
triangles. Prove that:
(i) PAB = QCD
(ii) PB = QD

Answer 1

Answer:

<1>

angle (PAD)=angle (QCB)=60° equilateral ∆ property(1)eq

angle (DAB)=angle(DCB) opposite angle of parallelogram (2)eq

adding eq 1&2

angle (PAD)+angle (DAB)=angle (QCB)+angle(DCB)

angle (PAB)=angle (QCD)

<11>

In∆PAB & ∆QCD

AB=DC(opposite side of parallelogram)

PA=DA=CB=QC then

PA=QC

angle (PAB)=angle (QCD)

∆PAB congruent to ∆QCD by(SAS)

PB=QD

Answer 2

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