ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at 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.


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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first you draw a figure then check out hte below solution :

ar(BDF) = ar (ADF)

becauz triangles on the same base DF and between same parallels BA and DC are = in area.

ar(BAF) = 1/2 ar(ABCD) ......(1)

Becauz triangle and parallelogram on the same base BA and between same parallels BA and CD are = in area.

ar (ADF) = ar (BCF) ....(2)

AS they have same altitude because between same parallels and same length of base DF = CF.It is proved below :

In triangle ADF and CFE


Angle CFE = Angle AFD (Vertically opposite angles)

angle CEF = =angle FAD (alternate interior angles)

so by ASA congruency rule

triangle ADF CONGRUENT triangle CFE

Therefore, CF = DF (BY C.P.C.T)

s, ar(ADF) + ar (BCF) + ar ( BFA ) = ar (ABCD)

2 ar(ADF) + 1/2 ar (ABCD) = ar (ABCD) {BY equation (1)} and {by equation (2)}

2 ar (ADF) = 1/2 ar(ABCD)

ar (ADF) = 1/2*2 ar (ABCD) = 1/4 ar (ABCD)

THEREFORE, ar (BDF) =1/4 ar (ABCD) { BY equation (2)}

hence proved

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