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Question
In quadrilateral ABCD, diagonals AC and BD intersect at point E such that
AE : EC = BE : ED. Show that: ABCD is a trapezium.
Solution
Given, AE : EC = BE : ED
Draw EF || AB
In ∆ABD, EF || AB
Using Basic Proportionality theorem,
`(DF)/(FA) = (DE)/(EB)`
But, given `(DE)/(EB) = (CE)/(EA)`
∴ `(DF)/(FA) = (CE)/(EA)`
Thus, in ∆DCA, E and F are points on CA and DA respectively such that
`(DF)/(FA) = (CE)/(EA)`
Thus, by converse of Basic proportionality theorem, FE || DC.
But, FE || AB.
Hence, AB || DC.
Thus, ABCD is a trapezium.
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