ABCD is a parellelogram and E is the mid point of BC.DE and AB on producing meet at F.Prove that AF=2AB
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IN TRIANGLE CDE AND BEF
AngleCED=AngleBEF (VOA)
Angle ECD=Angle EAF (Alternate)
BE=CE
So, triangle CDE Is congruent to BEF
BF=CD (Cpct)
BF=CD=AB(AB =CD)
AF=AB+BF
AF=AB+AB(AB=BF)
AF=2AB
I HOPE THAT IT HELPS ...
AngleCED=AngleBEF (VOA)
Angle ECD=Angle EAF (Alternate)
BE=CE
So, triangle CDE Is congruent to BEF
BF=CD (Cpct)
BF=CD=AB(AB =CD)
AF=AB+BF
AF=AB+AB(AB=BF)
AF=2AB
I HOPE THAT IT HELPS ...
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