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Question
In the given figure, if ABCD is a trapezium in which AB || CD || EF, then prove that `(AE)/(ED) = (BF)/(FC)`.
Solution
Given:- AB || CD || EF
To prove:- `(AE)/(ED) = (BF)/(FC)`
Constant:- Join BD to intersects EF at G.
Proof:- In ΔABD
EG || AB (EF || AB)
`(AE)/(ED) = (BG)/(GD)` (By BPT) ...(1)
In ΔDBC
GF || CD (EF || CD)
`(BF)/(FC) = (BG)/(GD)` (By BPT) ...(2)
From (1) and (2)
`(AE)/(ED) = (BF)/(FC)`
Hence Proved
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