The power of origami

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The power of origami

December 2009

Trisecting the angle using origami — proof

In the image on the right the angle between the blue line and the bottom edge is the angle to be trisected, and we must show that α=β=γ. The red line is the crease resulting from the fold in step 4 of the folding sequence. Now look at the triangle EBb. We know that the length of the line segment EG is equal to the length of the line segment GB, and we also know that the line segment Gb, which is the height of the triangle EBb, meets the line segment EB at a right angle. In other words, the height Gb of the triangle divides the opposite side into half. This means that the triangle EBb is isosceles. The mirror image of the triangle EBb when reflected in the red crease line is the triangle ebB, which is therefore also isosceles. The height of the triangle ebB extending from the point B therefore divides the angle at B into half. This shows that α=β. By mirror symmetry we have that β is equal to the angle δ of the triangle GbB. And since the line GH is parallel to the bottom edge BC, we have that γ=δ. This proves that β=γ.

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