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Question
XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that
ar (ABE) = ar (ACF)
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Solution
It is given that
XY || BC ⇒ EY || BC
BE || AC ⇒ BE || CY
Therefore, EBCY is a parallelogram.
It is given that
XY || BC ⇒ XF || BC
FC || AB ⇒ FC || XB
Therefore, BCFX is a parallelogram.
Parallelograms EBCY and BCFX are on the same base BC and between the same parallels BC and EF.
∴ Area (EBCY) = Area (BCFX) ... (1)
Consider parallelogram EBCY and ΔAEB
These lie on the same base BE and are between the same parallels BE and AC.
∴ Area (ΔABE) = 1/2Area (EBCY) ... (2)
Also, parallelogram BCFX and ΔACF are on the same base CF and between the same parallels CF and AB.
∴ Area (ΔACF) = 1/2Area (BCFX) ... (3)
From equations (1), (2), and (3), we obtain
Area (ΔABE) = Area (ΔACF)
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