ABCD is a parallelogram; E and f are the mid-points of AB and CD respectively.

Solution:

Since E and F are mid-points of AB and CD respectively

AE = BE = (1/2) AB

And CF = DF = (1/2) CD

But, AB = CD

(1/2)AB = (1/2) CD

⟹ BE = CF

Also, BE ∥ CF [∴ AB ∥ CD]

Therefore, BEFC is a parallelogram

BC ∥ EF and BE = PH  .... (i)

Now, BC ∥ EF

⟹ AD ∥ EF         [∵ BC ∥ AD as ABCD is a parallelogram]

Therefore, AEFD is a parallelogram.

⟹ AE = GP

But E is the mid-point of AB.

So, AE = BF

Therefore, GP = PH.