We can draw the trapezium as
Here, let EF be the line segment joining the oblique sides of the trapezium at midpoints E and F (say) correspondingly.
Construction: Extend AD and BC such that it meets at P.
To Prove: EF ∥ DC and EF ∥ AB
Proof: Given that, ABCD is trapezium which means DC ∥ AB. …(statement (i))
DC ∥ AB (by statement (i))
So, this means we can apply Thale’s theorem in ∆PAB. We get
∵ E and F are midpoints of AD and BC respectively, we can write
DA = DE + EA
Or DA = 2DE …(iii)
CB = CF + FB
Or CB = 2CF …(iv)
Substituting equation (iii) and (iv) in equation (ii), we get
By applying converse of Thale’s theorem, we can write DC ∥ EF.
Now if DC ∥ EF, and we already know that DC ∥ AB.
⇒ EF is also parallel to AB, that is, EF ∥ AB.
This means, DC ∥ EF ∥ AB.
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