41. 8. 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(ΔABE) = ar(ΔACF). 

Given: XY is a line parallel to side BC of a triangle ABC. BE || AC and CF || AB meet XY at E and F respectively.
To Prove: ar(ΔABE) = ar(ΔACF).
Proof: ∵ XY || BC    | given
and    CF || BX | ∵ CF || AB (given)
∴ □BCFX is a || gm
A quadrilateral is a parallelogram if its opposite sides are parallel



∴ BC = XF
| Opposite sides of a parallelogram are equal ⇒ BC = XY + YF    ...(1)
Again,
∵ XY || BC    | given
and    BE || CY    | ∵ BE || AC (given)
∴ □BCYE is a parallelogram
A quadrilateral is a parallelogram if its opposite sides are parallel
∴ BC = YE
| Opposite sides of a parallelogram are equal
⇒ BC = XY + XE    ...(2)
From (1) and (2),
XY + YF = XY + XE
⇒ YF = XE
⇒ XE = YF    ...(3)
∵    ΔAEX and ΔAYF have equal bases (∵ XE = YF) on the same line EF and have a common vertex A.
∴ Their altitudes are also the same.
∴ ar(ΔAEX) = ar(ΔAFY)    ...(4)
∵    ΔBEX and ΔCFY have equal bases (∵ XE = YF) on the same line EF and are between the same parallels EF and BC (∵ XY || BC).
∴ ar(ΔBEX) = ar(ΔCFY)    ...(5)
Two triangles on the same base (or equal bases) and between the same parallels are equal in area
Adding the corresponding sides of (4) and (5), we get
ar(ΔAEX) + ar(ΔBEX) = ar(ΔAFY) + ar(ΔCFY)
⇒    ar(ΔABE) = ar(⇒ACF).