Question:
ABCD is a parallelogram; E and f are the mid-points of AB and CD respectively. GH is any line intersecting AD, EF and BC at G, P and H respectively. Prove that GP = PH.
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.