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 ABE = ar ACF?

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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 ABE = ar ACF?

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8.XY is a line parallel to side BC of a triangle ABC. If BE || AC and ...

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).

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