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Question
E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF || AB and EF = `1/2` (AB + CD).
[Hint: Join BE and produce it to meet CD produced at G.]
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Solution
Given: ABCD is a trapezium in which AB || CD. Also, E and F are respectively the mid-points of sides AD and BC.
Construction: Join BE and produce it to meet CD produced at G, also draw BD which intersects EF at O.
To prove: EF || AB and EF = `1/2` (AB + CD).
Proof: In ΔGCB, E and F are respectively the mid-points of BG and BC, then by mid-point theorem,
EF || GC
But GC || AB or CD || AB ...[Given]
∴ EF || AB
In ΔADB, AB || EO and E is the mid-point of AD.
Therefore by converse of mid-point theorem, O is mid-point of BD.
Also, EO = `1/2` AB ...(i)
In ΔBDC, OF || CD and O is the mid-point of BD.
∴ OF = `1/2` CD [By converse of mid-point theorem] ...(ii)
On adding equations (i) and (ii), we get
EO + OF = `1/2` AB + `1/2` CD
⇒ EF = `1/2` (AB + CD)
Hence proved.
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