Given :
A trapezium ABCD in which AB∥DC and its diagonals AC and BD intersect at O.
Construction :
Through O, draw EO∥AB, meeting AD at E.
Proof :
In △ADC, EO∥DC
Therefore,
⇒= ……………..(1)
In △DAB,EO∥AB
Therefore,
⇒=..........(2)
From 1 and 2, we get,
⇒= Hence, We can say that The diagonals of a trapezium divide each other proportionally.
Consider second statement :
Here, ABCD is a trapezium such that EF||AB||DC
Join EF to cut AC at G.
ΔADC , EG||DC , so by basic proportionality theorem , we have
⇒
= ………………(1)
ΔACB, GF||AB , so by basic proportionality theorem , we have
⇒
=………………(2)
From (1) and (2)
⇒= Hence, we can say that Any line drawn parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally.
Hence, both the statements are true.