Exercise 10.3.2

Proof. (a) To bisect the same angle $𝜃$, we apply the same line $l_2$ on the bottom line. The fold is on the bisector of the angle. To double the angle, we fold the sheet along the line $l_2$. Then the bottom line goes on the line that makes a double angle with the bottom line. (b) If the angle $𝜃$ is between $0$ and $\pi ∕4$, the angle ${𝜃}^{\prime }=\pi ∕2-𝜃$ is constructible by origami with the symmetry with regard to the bisector of the bottom left corner, and ${𝜃}^{\prime }$ is between $\pi ∕4$ and $\pi ∕2$, so we can trisect it, and we obtain the angle $\pi ∕6-𝜃∕3$. As $\pi ∕6$ is origami-constructible by trisection of the angle $\pi ∕2$ (here $Q=P_2=Q_2$) and as we can add or subtract an origami-constructible angle, then so is the angle $𝜃∕3$.

If $𝜃$ is between $\pi ∕2$ and $\pi$, Then ${𝜃}^{″}=\pi -𝜃$ is origami-constructible and between $0$ and $\pi ∕2$, so we can trisect it. We obtain the angle $\pi ∕3-𝜃∕3$, where $\pi ∕3$ is the double of the origami-constructible angle $\pi ∕6$, so the angle $𝜃∕3$ is also origami-constructible. □