# Is Every Trapezoid Parallelogram?

Given trapezoid ABCD (AD||BC), extend BC to E so that

What if applet does not run? |

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

Denote

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

### References

- 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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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,

Apparently, we should not have divided by

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

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,

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 *strictly monotone increasing* which makes it impossible to have

## Related material
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## Geometric Fallacies | |

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Copyright © 1996-2018 Alexander Bogomolny

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