ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F.show that ar(triangle BDF)=1/4 ar(ABCD)

given: ABCD is a parallelogram. and BC=CE

TPT: are(ΔBDF)= 1/4 area(quadrilateral ABCD)

proof: since CF is parallel to AB,

ΔECF is similar to ΔEBA.

EB=EC+CB=EC+EC=2EC

the ratio of the sides will be equal .

therefore BF is the median of ΔBDC.

area( ΔBDF)=area( ΔBFC)

area( ΔBDF)=1/2area( ΔBDC)

=1/2(half of the area of parallelogram ABCD) [since diagonal of a parallelogram divides it in equal parts ]

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