ABCD is a parallelogram. E is the mid-point of the side BC, DE and AB when they meet at F. How can I prove that AF = 2AB?
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given = parallelogram ABCD with E as mid point of BC ie BE=CE
To prove - AF=2AB
In triangle(DEC) AND triangle(FEB)
angleECD = angleFBE ( alternate int angles)
CE=BE ( given)
angleDEC = angleFEB ( vertically opps angles)
triangle (DEC) ≈ triangle(FEB) {ASA}
DC = BF { cpct} - 1
AB=DC { given} - 2
from 1 and 2
AB = BF
2AB=2BF
2AB=AF
To prove - AF=2AB
In triangle(DEC) AND triangle(FEB)
angleECD = angleFBE ( alternate int angles)
CE=BE ( given)
angleDEC = angleFEB ( vertically opps angles)
triangle (DEC) ≈ triangle(FEB) {ASA}
DC = BF { cpct} - 1
AB=DC { given} - 2
from 1 and 2
AB = BF
2AB=2BF
2AB=AF
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