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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 area of △ABE=area of △ACF

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Solution

## ∵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=XFOpposite 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 getarea(△AEX)+area(△BEX)=area(△AFY)+area(△CFY)⇒ area(△ABE)=area(△ACF)Hence proved.

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