Question

in the adjoining in figure ABCD is a parallelogram in which Eand F are midpoints of ABand CD respectively if GH is a line segment that cuts AD and EF and BC at G, P and H respectively prove that GP = PH​

Answer 1

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

∴AE=BE=

2

1

AB and CF=DF=

2

1

CD

But, AB=CD

∴

2

1

AB=

2

1

CD⇒BE=CF

Also, BE∥CF [∵AB∥CD]

∴ BEFC is a parallelogram.

⇒BC∥EF and BE=PH ...(i)

Now, BC∥EF

⇒AD∥EF [∵BC∥AD as ABCD is a ∥

gm

]

⇒AEFD is a parallelogram

⇒AE=GP ...(ii)

But, E is the mid-point of AB.

∴AE=BE

⇒GP=PH [Using (i) and (ii)