finding area pls include the figure also

Asked by  | 15th Sep, 2008, 09:45: PM

Expert Answer:



Given : XY is a line parallel to side BC of  a triangle  ABC. If BE II AC and CF II AB meet XY at E and F respectively

To Prove :

ar(ABE) = ar(ACF)


XY // BC (Given )

EB // AC i.e EB // YC.

Quad EBCY is a parallelogram

Similarly FCBXis a parallelogram.

They are on the same base BC and between the same parallels BC and EF.

Therefore AR(//gm EBCY)  =  Ar (//gm FCBX)

Subtracting Area(Trap XYCB) from both sides , we get

Ar (triangle EBX) = Ar (triangle CFY)

Adding Ar (trangle AXY) to both sides , we have

ar(ABE) = ar(ACF)


Answered by  | 26th Sep, 2008, 05:38: PM

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