Question

Ad and Bc are the non-parallel sides of a trapezium ABCD and E is the mid-point of AD and EF || AB. the line segment EF meets BC in F . prove that F is the mid-point of BC

Answer 1

In fig
AB || CD
but, AB || EF ----- Given
•°• AB || EF || CD
By property of 'Intercepts Made By Three Parallel line
AE/ED=BF/FC ----(1)
But , AE=ED-----E is midpoint of AD
B putting AE=ED in equation 1
we get,
1=BF/FC
i.e. FC=BF
•°• F is midpoint of BC