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.

Answer 1

Since,

E and F are midpoints of AB and CD

∴ AE = BE = \(\frac{1}{2}\) AB

And,

CF = DF = \(\frac{1}{2}\) CD

∵ AB = CD

∴ \(\frac{1}{2}\)AB =  \(\frac{1}{2}\)CD

= BE = CF

∴ BEFC is a parallelogram 

= BE || EF and BF= PH -----(i)

Now, 

BC || EF

= AD || EF 

∵ BC || AD as ABCD is a parallelogram

= AEFD is a parallelogram

= AE = GP

But is the midpoint of AB

∴ AE = BE

= GP = PH proved.