Prove by vector method that the line segment joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides and equal to half of their difference.

Explanation

Let `square` ABCD is a trapezium in which `AB"||"DC.Pand Q` are the mid-points of the diagonals AC and BD respectively. Join DP and produced upto R.
`" "angle1=angle2`
`" "("alternate interior angles," AB"||"DC...(1)`

`In DeltaAPRand DeltaDPC,`
`because{:{(angle1,=angle2,["from"(1)]),(angle3,=angle4,("vertically opposite angles")),(AP,=CP,("given")):}`
`therefore" "DeltaAPR~=DeltaCPD" "("ASA theorem")`
`implies" "AR=CDand" "PR=DR" "(c.p.c.t.)...(2)`
`therefore` P is the mid-point of DR.
Now, in `DeltaDRB,`
Since P and Q are the mid-points of DR and DB respectively,
`therefore" "PQ"||"RBandPQ=1/2RB" "("mid-point theorem")`
`implies" "PQ"||"ABandPQ=1/2(AB-AR)`
`implies" "PQ"||"AB"||"CDandPQ=1/2(AB-CD)" "["from"(2)]`