Question

In quadrilateral ABCD, diagonals AC and BD intersect at point E such that
AE : EC = BE : ED. Show that: ABCD is a trapezium.

Answer 1

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.