Chasing Angles in Pascal's Hexagon

Pascal's Theorem: If the vertices of a hexagon lie on a circle and the three pairs of opposite sides intersect, then the three points of intersection are collinear. (Some intersections may lie at infinity which is the case where the corresponding sides are parallel.)

The applet below illustrates a simple proof due to Jan van Yzeren.

The vertices of the hexagon are numbered Ak, k = 0, 1, ..., 5, and A6 = A0. Assume Pk, k = 0, 1, 2 is the intersection of AkAk+1 and Ak+3Ak+4, k = 0, 1, 2. Consider the circle through A1, A4 and P1.

(In the applet the base circle can't be resized but can be moved by dragging the mouse.)

 

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?

The applet displays two sequences of equal angles that are either inscribed into one of the circles and are supported by the same arc or are complementary to such. Here's one sequence of related angles: B0B1A4, B0P1A4, B0A1A4, A4A5P2, A0A3A4. Look up the second one. This shows that triangles A0A3P2 and B0B1P1 have parallel sides, are similar and perspective at P0, the consequence of which is that points P0, P1, P2 are collinear. Q.E.D.

Point P0 may lie at infinity (when A0A1||A3A4. This does not affect the proof. One exceptional case is that of parahexagon where all points P lie at infinity and hence on the line at infinity.

References

  1. J. van Yzeren, A Simple Proof of Pascal's Hexagon Theorem<, The American Mathematical Monthly, Vol. 100, No. 10, (Dec., 1993), pp. 930-931

Pascal and Brianchon Theorems

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

Copyright © 1996-2017 Alexander Bogomolny

 62644334

Search by google: