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Question
Show that the line segment which joins the midpoints of the oblique sides of a trapezium is parallel sides
Solution
Let the trapezium be ABCD with E and F as the mid Points of AD and BC, Respectively Produce AD and BC to Meet at P.
In Δ PAB, DC || AB.
Applying Thales’ theorem, we get
`(PD)/(DA)=(PC)/(CB)`
Now, E and F are the midpoints of AD and BC, respectively
⇒ `(PD)/(2DE)=(PC)/(2CF)`
⟹ `(PD)/(DE)=(PC)/(CF)`
Applying the converse of Thales’ theorem in Δ PEF, we get that DC
Hence, EF || AB.
Thus. EF is parallel to both AB and DC.
This completes the proof.
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