Is Every Trapezoid Parallelogram?

Given trapezoid ABCD (AD||BC), extend BC to E so that CE = AD and AD to F so that FA = BC.


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Consider the following derivation [Movshovitz-Hadar, pp. 70-71]:

Denote FA = BC = b and AD = CE = a. Let also x = AG, y = GH, z = CH.There are several pairs of similar triangles.

Triangles CBG and ADG are similar, implying a proportion

b / a = CG / AG = (y + z) / x.

In other words,

(1) x = (a / b) (y + z).

Triangles AFH and CEH are similar, implying

b / a = AH / CH = (x + y) / z.

In other words,

(2) z = (a / b) (x + y).

Subtracting (1) from (2) we obtain

z - x = (a / b) (x - z).

Dividing by (z - x) and taking absolute values yields |a|/|b| = 1, that is, |a| = |b|. But this is only possible if ABCD is a parallelogram. Is there a paradox, or what?

Explanation

References

  1. N. Movshovitz-Hadar, J. Webb, One Equals Zero and Other Mathematical Surprises: Paradoxes, Fallacies, Mind Booglers , Key Curriculum Press (December 15, 1998)

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This is of course not true that all trapezoids are parallelograms. (In fact, as a matter of definition, none of them may be. According to one of the definitions, trapezoid is a quadrilateral with two parallel and unequal sides.)

So where did we cheat? By taking the absolute value of the meaningless identity, a / b = -1. Why is it meaningless? Because both a and b were introduced as length of the two bases of a trapezoid. So naturally both had to be positive. Then how come we got that identity? Ah, this is where an error has been committed!

Apparently, we should not have divided by (z - x). And there is only one reason why one should not divide by an expression, viz., when the expression is zero. Thus, indirectly we are led to a correct conclusion: x = z. AG = CH.


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More directly we arrive at the same conclusion by considering other pairs of similar triangles. Since FA and BC are parallel and equal, the quadrilateral AFBC is a parallelogram. ACED is another one. Hence BF||AC. ΔBDF is similar to ΔGDA so that

x / BF = a / (a + b).

In other words,

(3) x = BF × a / (a + b).

Triangles BEF and CEH are similar, i.e.

z / BF = a / (a + b),

or,

(4) z = BF × a / (a + b).

Comparing (3) and (4) we conclude that indeed x = z.

There are other ways to derive that identity. For example (W. McWorter), observe that ACED is also a parallelogram so that DE||AC||BF. Thus, by Euclid I.37, triangles BDF and BEF have the same area. Triangles GDE and CEH constitute the same fraction of triangles BDF and CEH, respectively; so they, too, have equal areas. By the previous argument, they have equal altitudes to the bases AD and CE which shows that the latter are equal, AD = CE.

And of course the same result could be obtained algebraically from (1) and (2). Indeed, the two imply that

(x + y) / z = (z + y) / x.

Rewrite this as x² + xy = z² + zy, or, with function f(w) = w² + wy, f(x) = f(z). But f is a parabola with the lowest point at (-y/2, -y²/4), which says that, for w > -y/2 and, certainly, for positive w, f(w) is strictly monotone increasing which makes it impossible to have f(x) = f(z), for x ≠ z when both are positive.

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  • All Triangles Are Isosceles
  • Two Perpendiculars From a Point to a Line
  • Every Parallelogram Is a Rectangle
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