## 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 _{6} = A_{0}._{k}, k = 0, 1, 2 is the intersection of A_{k}A_{k+1} and A_{k+3}A_{k+4}, k = 0, 1, 2. Consider the circle through A_{1}, A_{4} and P_{1}.

(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: B_{0}B_{1}A_{4}, B_{0}P_{1}A_{4}, B_{0}A_{1}A_{4}, A_{4}A_{5}P_{2}, A_{0}A_{3}A_{4}. Look up the second one. This shows that triangles A_{0}A_{3}P_{2} and B_{0}B_{1}P_{1} have parallel sides, are similar and perspective at P_{0}, the consequence of which is that points P_{0}, P_{1}, P_{2} are collinear. Q.E.D.

Point P_{0} may lie at infinity (when A_{0}A_{1}||A_{3}A_{4}. 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

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