Question

A point E is taken on the side CD of a parallelogram ABCD and CD is produced to F, so that DF=CE. BE produced meets AD produced in G and the line through F parallel to AG in H. Prove that parallelogram AFGH= parallelogram ABCD.​
it's very urgent pls solve

Answer 1

Answer:

⇒ QD=CP [ Given ] ---- ( 1 )

⇒ AB=CD [ Opposite sides of parallelogram ]

⇒ AB=CP+DP

⇒ AB=QD+DP [ From ( 1 ) ]

⇒ AB=QP

∴ AB∥QP [ Since, AB∥CD ]

∴ ABPQ is a parallelogram.

∴ AQ∥BP

As SR is extended part of BP

⇒ AQ∥SR

⇒ AQ∥RP

⇒ QS∥AR

∴ ARSQ is a parallelogram.