Prove the segments joining the midpoint of consecutive sides of an isosceles trapezoid form a rhombus.

Find the coordinates of midpoint D.

(–2a – 2b, c)

(2a – 2b, –c)

(–a + b, –c)

(–a – b, c)

95141 1404 393

  D  (-a-b, c)

Step-by-step explanation:

The midpoints of the sides are ...

  D = (T +R)/2 = ((-2a, 0) +(-2b, 2c))/2 = (-2a-2b, 2c)/2 = (-a-b, c)

  E = (R +A)/2 = ((-2b, 2c) +(2b, 2c))/2 = (0, 4c)/2 = (0, 2c)

  F = (A +P)/2 = ((2b, 2c) +(2a, 0))/2 = (2a +2b, 2c)/2 = (a+b, c)

  G = (P +T)/2 = ((2a, 0) +(-2a, 0))/2 = (0, 0)/2 = (0, 0)

(a) Point D is (-a-b, c) . . . . . matches the last choice

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(b) DEFG will be a rhombus if its diagonals are perpendicular and if their midpoints are the same.

  Midpoint of DF = (D +F)/2 = ((-a-b, c) +(a +b, c))/2 = (0, 2c)/2 = (0, c)

  Midpoint of EG = (E +G)/2 = ((0, 2c) +(0, 0))/2 = (0, 2c)/2 = (0, c)

The midpoints are the same, so the diagonals bisect each other. The figure is a parallelogram.

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The diagonals will be at right angles if the the "dot product" of their directions is zero. The "direction" can be found as the difference of the end-point coordinates:

  DF = F - D = (a +b, c) -(-a -b, c)) = (2a +2b, 0)

  EG = G -E = (0, 0) -(0, 2c) = (0, -2c)

The "dot product" is the sum of the products of the corresponding coordinates:

  (x1, y1) · (x2, y2) = x1×x2 +y1×y2

  DF · EG = (2a +2b, 0) · (0, -2c) = (2a +2b)×0 + 0×(-2c) = 0 +0 = 0

The diagonals are perpendicular bisectors of each other, so the figure DEFG is a rhombus.

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