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# Geometry Trouble!

Please provide a solution to this problem.

$$ABCD$$ is a trapezium with $$AD$$ parallel to $$BC$$; $$AD=3BC$$ and a transversal $$XY$$ cuts $$BC$$ at $$X$$ and $$AD$$ at $$Y$$. If $$EF$$ is a line segment contained in $$XY$$ such that $$AE$$ is parallel $$DF$$. $$BE$$ parallel to $$CF$$ and $$AE/DF=CF/BE=2$$, show that area of the triangle $$EFD$$ is equal to $$\dfrac14$$ of the area of $$ABCD$$.

Note by Achal Jain
5 months ago

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@achal jain bhai is this problem from some book.. · 4 months, 2 weeks ago

yeah from "Challenges of Pre College Mathematics". · 4 months, 2 weeks ago