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Question
Prove that the line segment joining the mid-points of the diagonals of a trapezium is parallel to each of the parallel sides, and is equal to half the difference of these sides.
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Solution
Join AC and BD. M and N are mid-points of AC and BD respectively. Join MN. Draw a line CN cutting AB at E.
Now, in Δs DBC and BNE,
DN = NB ...(N is the mid-point of BD, given)
∠CDB = ∠EBN ...(Alternate angles as DC || AB)
∠DNC = BNE ...(Vertically opposite angles)
⇒ ΔDNC ≅ ΔBNE ...(By A-S-A Test)
⇒ DC = BE
By Mid-point Theorem, in ΔACE, M and N are mid-points
MN = `(1)/(2)"AE" and "MN" || "AE" or "MN" || "AB"`
Also, AB || CD, therefore, MN || CD
⇒ MN = `(1)/(2)["AB" = "BE"]`
⇒ MN = `(1)/(2)["AB" = "CD"]` ...(since BE = CD)
⇒ MN = `(1)/(2)` x Difference of parallel sides AB and CD.
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