In a trapezium ABCD, AB is parallel to CD and DC=2AB. A line EF drawn parallel to AB cuts AD in F and BC in E such that BE/EC=3/4. diagonals DB intersects EF at G prove 7FE =10 AB
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Step-by-step explanation:
See the diagram attached with the data in the question.
AB = a
Given that CD = 2AB = 2a.
If a line BI is drawn from B to CD, parallel to AD, it will make right angle at BIC.
Since AB = DI, IC = 2a – a = a.
In ΔBIC and ΔBHE,
∠HBE = ∠IBC
∠BHE = ∠BIC
∠ICB = ∠HEB (HE is parallel to IC and EC is the tangent cutting both parallel lines.)
ΔBIC ~ ΔBHE
BE/BC = HE/IC
3x/7x = HE/a
HE = 3a/7
EF = FH + HE = a + 3a/7 = 10a/7
7EF = 10a = 10AB
Hence proved.
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