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Prove that the straight line joining the midpoints of the diagonals of a trapezium is parallel to the parallel sides and is equal to half their difference.

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ABCD is a trapezium with AB parallel to CD and midpoints of AC and BD be P and Q
repectively........
CONSTRUCTION - Draw a line from D passing through Pand meeting AB in E............PROOF:
In triangle APE and triangle CPD
angle PAE = angle PCD ( alternate angles )
Angle PEA = angle PDC ( alternate angles )
AP = PC ( P is the midpoint of AC )
triangle APE is congruent to triangle CPD
PE = PD ( CPCT )
In triangle BED
P is the midpoint of DE ( proved above )
Q is the midpoint of BD ( given )
Therefore PQ is parallel to BE ( midpoint theorem )
SO PQ is parallel to AB
AB parralel to CD ( given )
SO PQ is parallel to CD and AB
And one more step is...Join DE and produce to meet AB at G
In triangle AEG and CEG
<1 = <2 (Alternate interior angles)
AE = EC (E is the mid point of AC)
<4 = <3 ( Vertically opposite angles)
Triangle AEG is congruent to triangle CED (by ASA congruence rule)
DE = EG ( CPCT)
AG = CD ( CPCT)
In triangle BGD
E is the mid point of DG
F is the mid point of BD
By mid point theorem,
EF II BG
EF II AB
AB II CD
Hence, AB II EF II CD
EF = 1/2 BG
EF = 1/2 ( AB - AG)
EF = 1/2 (AB - CD)
HOPE THIS HELPS :)