Question

In a trapezium ABCD, AB is parallel to DC and DC=3AB . EF is parallel to AB intersects DA and CB at E and F such that BF/FC=2/3. Prove that 3DC= 5EF.

Answer 1

[tex] \frac{OF}{3y} = \frac{2x}{5x} OF = \frac{6y}{5} is the answer .................[/tex] ..............

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