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 A_k, k = 0, 1, ..., 5, and A₆ = A₀. Assume P_k, k = 0, 1, 2 is the intersection of A_kA_k+1 and A_k+3A_k+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

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

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