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ABCD is a parallelogram. AB is produced ...

`ABCD` is a parallelogram. `AB` is produced to `E` so that `B E=A B`. Prove that `ED` bisects `BC`

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"here`AB=BE` (given) `AB=CD` (opposite sides of a parallelogram)
therefore `BE=CD`....(1) Let DE intersects BC at F Now in `/_\CDF` and` /_\BEF` `/_DCF=/_EBF ` (since `AB∥CD`)
`/_DFC=/_EFB` (vertically opposite angles)
`BE=CD` (from (1))
from angle sum `/_CDF=/_BEF`
...
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