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 Proved