CBSE Class 9 Answered
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)
Proof:
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)
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