Question

In the given figure ABCD is a parallelogram in which BC is produced to such that CE=BC. AE intersects CD at F. Show that at(BDF)=1/4 ar(ABCD) please answer

Answer 1

first you draw a figure then check out the 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

CE = AD (Becauz CE = BC = AD SIDES OF PARALLELOGRAM )

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