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In a parallelogram `ABCD ,` the bisector of `/_A` also bisects `BC` at `X` Prove that `A D=2A B` ​

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"Let The bisector of `/_A` bisects the side BC at X.
given ABCD is a parallelogram so `AD∥BC`
now, AX is the transversal so,` /_1=/_3`(alternate angles).....(1)
and`/_1=/_2`(AX is the bisector of` /_A`)....(2)
From(1)and(2) we get
...
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