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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 :)
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