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A point E is taken as the midpoint of the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that ar (ADF) = ar(ABFC).

Asked by Topperlearning User 4th June 2014, 1:23 PM
Answered by Expert

In triangles ABE and CFE,

BE = CE (E is the mid-point of BC)

(vertically opposite angles)

(AB||CD and CD is produced to CF)

ABE FCE (ASA criteria)


Now, AB = CD, as ABCD is a parallelogram.

Therefore, AB = CF = CD

C is midpoint of FD

In parallelogram ABFC, AF is the diagonal, therefore,

ar(ABF) = ar(ACF) ……(i)


AC is the median, since CD = CF

Therefore, ar(ACF) = ar(ACD) ……(ii)

From (i) and (ii)

ar(ABF) = ar(ACD)

Adding ar(ACF) to both sides,

ar(ABF) + ar(ACF) = ar(ACD) + ar(ACF)

ar(||gm ABFC) = ar(ADF)

Answered by Expert 4th June 2014, 3:23 PM
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