If P is a point and ABCD is a quadrilateral and
show that ABCD is a parallelogram.
Given a quadrilateral ABCD, P is a point outside the quadrilateral and ![]()

[given]
Or, ![]()
………..(i)[as
]
Consider ΔAPB and apply triangle law of vector, we get
![]()
And consider ΔDPC and apply triangle law of vector, we get
![]()
Substitute the values from eqn(ii) an eqn(iii) in eqn(i), we get
![]()
Therefore, AB is parallel to DC and equal is magnitude.
Hence, ABCD is a parallelogram.
Hence proved
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