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
(In the applet the base circle can't be resized but can be moved by dragging the mouse.)
|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.
- J. van Yzeren, A Simple Proof of Pascal's Hexagon Theorem<, The American Mathematical Monthly, Vol. 100, No. 10, (Dec., 1993), pp. 930-931
Copyright © 1996-2018 Alexander Bogomolny