Question

ABCD is a parallelogram, E is the midpoint of AB and F is the mid-point of CD. PQ is any line that intersects AD, EF and BC at P, G and Q. Prove that PG = GQ.​

Answer 1

Answer:

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)]

Step-by-step explanation:

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