Two Perpendiculars From a Point to a Line
How Is It Possible?


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What if applet does not run?

Two circles are constructed on two segments AB and AC as diameters. The line BC intersects one of the circles in D and the other in E. Both angles AEC and ADB are right as inscribed angles subtending a diameter. It appears like there are two perpendiculars AD and AE from A to BC. How can this be?

But there is something else going on. Besides A, the circles have another common point, say, K. The angles AKB and AKC also subtend diameters of the circles. Therefore, KC is a continuation of the line KB. It appears that there are two straight lines - the original AB and the constructed AKB - that pass through two points B and C. Something is indeed wrong. But what?

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Copyright © 1996-2018 Alexander Bogomolny


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?

One solution to two fallacies: the three points K, D, and E are one and the same. Indeed, it must be clear that K lies on BC. (The applet cheats by drawing circles a little displaced from the declared centers.)

References

  1. V. M. Bradis et al, Lapses in Mathematical Reasoning, Dover, 1999, pp. 138-139

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  • A Circle With Two Centers
  • Rouse Ball's Fallacy
  • All Triangles Are Isosceles
  • Is Every Trapezoid Parallelogram?
  • Every Parallelogram Is a Rectangle
  • |Activities| |Contact| |Front page| |Contents| |Geometry| |Fallacies|

    Copyright © 1996-2018 Alexander Bogomolny

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