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Vector Algebra

Question
CBSEENMA12034045

Prove, using vectors, that the line segment joining the mid-points of the non-parallel sides of a trapezium is parallel to the bases and is equal to half the sum of their lenghts.

Solution

Let straight a with rightwards arrow on top comma space straight b with rightwards arrow on top comma space straight c with rightwards arrow on top comma space straight d with rightwards arrow on top be the position vectors of the vertices A, B, C, D of the trapezium in which AB||CD
Now AB || CD
  rightwards double arrow space space AB with rightwards arrow on top space equals space straight lambda space DC with rightwards arrow on top
where straight lambda is some scalar.
  therefore space space space space straight P. straight V. space of space straight B space minus space straight P. straight V. space of space straight A space equals space space straight lambda space left parenthesis straight P. straight V. space of space straight C space minus space straight P. straight V. space of space straight D right parenthesis
rightwards double arrow space space space space straight b with rightwards arrow on top space minus space straight a with rightwards arrow on top space equals straight lambda left parenthesis straight c with rightwards arrow on top space minus space straight d with rightwards arrow on top right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis

Let E be mid-point of AD and F be mid-point of BC.
therefore space space space straight P. straight V. space of space straight E space equals space fraction numerator straight a with rightwards arrow on top plus straight d with rightwards arrow on top over denominator 2 end fraction space and space straight P. straight V. space of space straight F space equals fraction numerator straight b with rightwards arrow on top space plus straight c with rightwards arrow on top over denominator 2 end fraction
therefore space space space space space EF with rightwards arrow on top space equals space straight P. straight V. space of space straight F space minus space straight P. straight V. space of space straight E space equals space fraction numerator straight b with rightwards arrow on top space plus space straight c with rightwards arrow on top over denominator 2 end fraction space minus space fraction numerator straight a with rightwards arrow on top plus straight d with rightwards arrow on top over denominator 2 end fraction
space space space space space space space space space space space space space space equals space fraction numerator straight b with rightwards arrow on top plus straight c with rightwards arrow on top minus straight a with rightwards arrow on top minus straight d with rightwards arrow on top over denominator 2 end fraction space equals fraction numerator open parentheses straight b with rightwards arrow on top minus straight a with rightwards arrow on top close parentheses space plus left parenthesis straight c with rightwards arrow on top minus straight d with rightwards arrow on top right parenthesis over denominator 2 end fraction space equals fraction numerator straight lambda space left parenthesis straight c with rightwards arrow on top minus straight d with rightwards arrow on top right parenthesis space plus space left parenthesis straight c with rightwards arrow on top minus straight d with rightwards arrow on top right parenthesis over denominator 2 end fraction
space space space space space space space space space space space space space space equals space open parentheses fraction numerator straight lambda plus 1 over denominator 2 end fraction close parentheses space space left parenthesis straight c with rightwards arrow on top space minus space straight d with rightwards arrow on top right parenthesis space equals space fraction numerator straight lambda plus 1 over denominator 2 end fraction DC with rightwards arrow on top
∴     EF and DC are parallel.
Also EF is parallel to AB as AB is parallel to DC.
Now, EF space equals space fraction numerator straight lambda plus 1 over denominator 2 end fraction DC space equals fraction numerator straight lambda. space DC space plus DC over denominator 2 end fraction space rightwards double arrow space space space EF space equals space fraction numerator AB plus DC over denominator 2 end fraction
Hence the result.