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.

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

What if applet does not run?

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)

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

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.

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

What if applet does not run?

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.

Related material
Read more...

Geometric Fallacies

  • A Circle With Two Centers
  • Rouse Ball's Fallacy
  • All Triangles Are Isosceles
  • Two Perpendiculars From a Point to a Line
  • Every Parallelogram Is a Rectangle

    • |Activities| |Contact| |Front page| |Contents| |Geometry|

    Copyright © 1996-2018 Alexander Bogomolny

    72083917