Let AB, CD be two line segments such that AB ║ CD and AD║BC. Let E be the midpoint of BC and let DE extended meet AB in F. Prove that AB = BF.
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Given : AB//CD , AD//BC
ABCD is a parallelogram.
E is the mid point of BC .
Let G be the mid point of AD .
Join G , E .
Now In ∆AFD , GE is the line
joining the mid point F , E of two
sides AD and FD.
Therefore ,
GE // AF and
GE = AF/2
=> AB = AF/2
[ GE = AB, Opposite sides of
parallelogram ABEG ]
=> 2AB = AF ----- ( 1 )
Now ,
AF = AB + BF
=> 2AB = AB + BF [ from ( 1 ) ]
=> AB = BF
••••••
ABCD is a parallelogram.
E is the mid point of BC .
Let G be the mid point of AD .
Join G , E .
Now In ∆AFD , GE is the line
joining the mid point F , E of two
sides AD and FD.
Therefore ,
GE // AF and
GE = AF/2
=> AB = AF/2
[ GE = AB, Opposite sides of
parallelogram ABEG ]
=> 2AB = AF ----- ( 1 )
Now ,
AF = AB + BF
=> 2AB = AB + BF [ from ( 1 ) ]
=> AB = BF
••••••
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