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 ]