ABCD is a parallelogram , a line trhough A cuts DC at oint F and BC produced at E Prove that &#x2206 BCF is equal in area to &#x2206 DFE.

Construction: AC is joined.

AB || CD and AD || BC .
Now &#x2206 ACE and &#x2206 DCE are on the same base CE and between sane parallel lines AD and AE . Therefore
&#x2206 ACE = &#x2206 DCE
&#x2206 ACE – &#x2206 FCE = &#x2206 DCE – &#x2206 FCE
∴ &#x2206 ACF = &#x2206 DFE
Now &#x2206 ACF and &#x2206 BCF are on the same base and between same parallel .
∴ &#x2206 ACF = &#x2206 BCF
∴ &#x2206 DFE = &#x2206 BCF

Proved

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