Question

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)
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Answer 1

Answer:

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Answer 2

Answer:

∵XY∥BC(given)

and CF∥BX(given) ∵CF∥AB

∴BCFX is a parallelogram.

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)

∴BCYE is a parallelogram.

A quadrilateral is a parallelogram if its opposite sides are parallel

∴BC=YE

⇒BC=XY+YE ...(2)

From (1) and (2),

XY+YF=XY+XE

⇒YF=XE ...(3)

∵△AEX and △AYF have equal bases (∵XE=YF) on the same line EF and have a common vertex A

∴ Their altitudes are also the same.

∴area(△AEX)=area(△AFY) ....(4)

∵△BEX and △CFY have equal bases (∵XE=YF) on the same line EF and between the same parallels EF and BC$

∴ area(△BEX)=area(△CFY) ....(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

area(△AEX)+area(△BEX)=area(△AFY)+area(△CFY)

⇒ area(△ABE)=area(△ACF)

Hence proved.