Exercise 10.1.2

Proof.

(a)

First we construct a triangle ${\alpha }^{\prime },{\beta }^{\prime },{\gamma }^{\prime }$ isometric to $\alpha ,\beta ,\gamma$, where ${\alpha }^{\prime }=0$. The circle $C(0,r)$, with $r=||\stackrel{\to }{\mathit{\alpha \beta }}||$ is constructible (rule C2). The intersection of $C(0,r)$ with the constructible $x$-axis contains two constructible points (rule P2), one of them, named ${\beta }^{\prime }$, with positive $x$-coordinate.

The circles $C({\alpha }^{\prime },r^{\prime })=C(0,r^{\prime })$ and $C({\beta }^{\prime },r^{″})$ where $r^{\prime }=||\stackrel{\to }{\mathit{\alpha \gamma }}||,r^{″}=||\stackrel{\to }{\mathit{\beta \gamma }}||$ are constructible (rule C2). Since

$$||\stackrel{\to }{{\alpha }^{\prime }{\beta }^{\prime }}||=r=||\stackrel{\to }{\mathit{\alpha \beta }}||<||\stackrel{\to }{\mathit{\alpha \gamma }}||+||\stackrel{\to }{\mathit{\beta \gamma }}||=r^{\prime }+r^{″},$$

the intersection $C({\alpha }^{\prime },r^{\prime })\cap C({\beta }^{\prime },r^{″})$ is nonempty, so $C({\alpha }^{\prime },r^{\prime })\cap C({\beta }^{\prime },r^{″})=\{{\gamma }^{\prime },{\gamma }^{″}\}$, where ${\gamma }^{\prime }$ has positive $y$-coordinate, and ${\gamma }^{\prime }$ is constructible by rule $P3$.

Since $||\stackrel{\to }{\mathit{\alpha \beta }}||=||\stackrel{\to }{{\alpha }^{\prime }{\beta }^{\prime }}||,||\stackrel{\to }{\beta ,\gamma }||=|\stackrel{\to }{{\beta }^{\prime },{\gamma }^{\prime }}||,||\stackrel{\to }{\mathit{\alpha \gamma }}||=||\stackrel{\to }{{\alpha }^{\prime }{\gamma }^{\prime }}||$, the two triangles $(\alpha ,\beta ,\gamma )$ and ( ${\alpha }^{\prime },{\beta }^{\prime },{\gamma }^{\prime })$ are isometric. Therefore the corresponding angles are equal :

$$\widehat{(\stackrel{\to }{\mathit{\alpha \beta }},\stackrel{\to }{\mathit{\alpha \gamma }})}=\pm \widehat{(\stackrel{\to }{{\alpha }^{\prime }{\beta }^{\prime }},\stackrel{\to }{{\alpha }^{\prime }{\gamma }^{\prime }})}.$$

If we take $\delta ={\gamma }^{\prime }$, then the angle $\widehat{(\stackrel{\to }{\mathit{\alpha \beta }},\stackrel{\to }{\mathit{\alpha \gamma }})}$ formed by $\alpha ,\beta ,\gamma$ is congruent to the angle formed by $0,1,\delta$.

(b)

$\text{∙}$ If ${\zeta }_n$ is constructible, so are $1,{\zeta }_n,\dots ,{\zeta }_n^{n-1}$ since $𝒞$ is a subfield of $ℂ$ (Theorem 10.1.4), and these points are the vertices of a regular $n$-gon.

$\text{∙}$ Suppose that a regular $n$-gon can be constructed by straightedge and compass. Let $\beta ,\gamma$ be two consecutive vertices. Then the center $\alpha$ of the $n$-gon is constructible, and the measure of the angle formed by $\alpha ,\beta ,\gamma$ has measure $𝜃=2\pi ∕n$ (see Example 10.1.3).

By part (a), we can construct $\delta$ with positive $y$-coordinate such that the angle formed by $0,1,\delta$ has the same measure $2\pi ∕n$. The intersection $\zeta$ of the line $0\delta$ with $C(0,1)$ is constructible, and $\arg <!--\; FUNCTION\; APPLICATION\; -->(\zeta )=2\pi ∕n,|\zeta |=1$, so ${\zeta }_n=\zeta$ is constructible.