Prove that the line segment joining the midpoints of the diagonals of a trapezium is parallel to each of the parallel sides and is equal to half of the difference of these sides.


PQ || AB, PQ =  

|AB - CD|

Step by Step Explanation:
  1. Let us join CQ and produce to meet AB at X. Now consider the triangle CDQ and BQX:
    Angle DCQ = Angle XBQ (Alternate interior angles)
    Angle DQC = Angle BQX (Vertically opposite angles)
    DQ = BQ (Q id midpoint of DB)
    Triangles CDQ and BQX are congruent.
  2. From step 2 we have QX = CQ. Now consider triangles CXA and CPQ. P and Q are midpoints of CA and CX, which means PQ is parallel to AX. Since PQ is parallel to AX or AB, it can be said that PQ is parallel to CD.
  3. Since, P and Q are mid points of CD and AB, it can be said that PQ =  
      × AX
    Or, PQ =  
      × (AB - BX)
    Since, we proved above that BX = CD, PQ = PQ || AB, PQ =  
    |AB - CD|

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