ABCD is a parallelogram and E is the midpoint of side BC . If DE and AB when product meet at F . prove that AF = 2Ab
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ABCD is a parallelogram
E is the mid point of BC,
DE and AB produced to F
To prove:AF=2AB
In ∆BEF,∆CDE
<BEF=<ECD
BC=BC
<DCE=<EBF
∆BEF=~∆CDE(byS.A.S rule)
BF=CCD(byCPCT rule)
AB=CD
AF=AB+BF(BF=CD)
AF=AB+AB(AB=CD)
AB=2AB
E is the mid point of BC,
DE and AB produced to F
To prove:AF=2AB
In ∆BEF,∆CDE
<BEF=<ECD
BC=BC
<DCE=<EBF
∆BEF=~∆CDE(byS.A.S rule)
BF=CCD(byCPCT rule)
AB=CD
AF=AB+BF(BF=CD)
AF=AB+AB(AB=CD)
AB=2AB
vivekTalla1:
last line mistake sorry correct :AF=2AB
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