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 $\alpha = \beta = \gamma$ .

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 $\alpha = \beta$ . By mirror symmetry we have that $\beta$ is equal to the angle $\delta$ of the triangle $GbB$ . And since the line $GH$ is parallel to the bottom edge $BC$ , we have that $\gamma = \delta$ . This proves that $\beta = \gamma$ .

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