ABCD is a parallelogram , a line trhough A cuts DC at oint F and BC produced at E Prove that ∆ BCF is equal in area to ∆ DFE.

# Construction: AC is joined.

AB || CD and AD || BC .

Now ∆ ACE and ∆ DCE are on the same base CE and between sane parallel lines AD and AE . Therefore

∆ ACE = ∆ DCE

∆ ACE – ∆ FCE = ∆ DCE – ∆ FCE

∴ ∆ ACF = ∆ DFE

Now ∆ ACF and ∆ BCF are on the same base and between same parallel .

∴ ∆ ACF = ∆ BCF

∴ ∆ DFE = ∆ BCF