Question

ABCD is a parallelogram ,E and F are midpoints of AB and CD respectively. GH is any line intersecting AD,EFand BC at G,P and H respectively. prove that GP=PH

Answer 1

Given: ABCD is a parallelogram, E and F are mid points of AB and DC and a line GH intersects AD, EF and BC at G, P, and H respectively.



Construction: Join GB and Let it intersect EF of X.



Now since E and F are mid points of AB and DC

⇒ AE = EB = AB
and DF= FC = DC
but AB = DC
⇒ AEFD and BCFE are parallelogram
⇒ AD EF BC ........ (1)

Now In ∆ ABG
AE = EB and EX AG (from (1))
⇒ X is the mid point of GB (mid point theorem)

and in ∆ GBH
GX = XB (∵ X is the mid point of GB)
and XP BH (from (1))
⇒ P is the mid point of GH
⇒ GP = PH

Answer 2

Given: ABCD is a parallelogram, E and F are mid points of AB and DC and a line GH intersects AD, EF and BC at G, P, and H respectively.

Construction: Join GB and Let it intersect EF of X.

Now since E and F are mid points of AB and DC

⇒ AE = EB = 1/2 AB

and DF= FC = 1/2 DC

but AB = DC

⇒ AEFD and BCFE are parallelogram

⇒ AD ll EF ll BC        ........ (1)

Now In ∆ ABG

AE = EB and EX ll AG               (from (1))

⇒ X is the mid point of GB                (mid point theorem)

and in ∆ GBH

GX = XB                (∵ X is the mid point of GB)

and XP ll BH               (from (1))

⇒ P is the mid point of GH

⇒ GP = PH

                                   Thus proved !!!