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Question
In trapezium ABCD, AB || DC, E and F are points on non-parallel sides AD and BC respectively, such that EF || AB. Show that = `"AE"/"ED" = "BF"/"FC"`
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Solution
Given: ABCD is a trapezium AB || DC
E and F are the points on the side of AD and BC
EF || AB
To Prove: `"AE"/"ED" = "BF"/"FC"`
Construction: Join AC intersecting AC at P
Proof:
In ∆ABC, PF || AB ...(Given)
By basic proportionality theorem
`"AP"/"PC" = "BF"/"FC"` ...(1)
In the ∆ACD, PE || CD ...(Given)
By basic Proportionality theorem
`"AP"/"PC" = "AE"/"ED"` ...(2)
From (1) and (2) we get
`"AE"/"ED" = "BF"/"FC"`
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