Exercise 10.3.1

Proof.

(a)

The dashed line $l$ is the fold of the sheet that applies $P_1$ on $Q_1$ and $P_2$ on $Q_2$. Let $s$ be the reflection (orthogonal symmetry) with regard to the line $l$.

Then $s(P_2)=Q_2$, and $s(Q_1)=P_1$, so $s$ applies the line $\overline{P_2{Q}_1}$ on the line $\overline{P_1{Q}_2}$. Let $Q$ be the intersection point of these two lines.

Then $s(Q)$ is on the images of these two lines, so

$$s(Q)\in s(\overline{P_1{Q}_2})\cap s(\overline{P_2{Q}_1})=\overline{P_2{Q}_1}\cap \overline{P_1{Q}_2}=\{Q\},$$

therefore $s(Q)=Q$.

Since the set of points $M$ such that $s(M)=M$ is the line $l$, we conclude that $Q\in l$.

(b)

Let $D$ the point at the bottom right corner, and $\delta$ a measure of the angle $(\widehat{\stackrel{\to }{P_{1}D},\stackrel{\to }{P_1,Q_1}})$.

As $(\widehat{\stackrel{\to }{P_{1}D},\stackrel{\to }{P_1{Q}_1}})+(\widehat{\stackrel{\to }{P_1{Q}_1},\stackrel{\to }{P_1{Q}_2}})=(\widehat{\stackrel{\to }{P_{1}D},\stackrel{\to }{P_1{Q}_2}})$, then $𝜃=\delta +\alpha$.

Since $P{Q}_1$ and $P_{1}D$ are parallel, $(\widehat{\stackrel{\to }{P_{1}D},\stackrel{\to }{P_1{Q}_1}})=(\widehat{\stackrel{\to }{Q_{1}P},\stackrel{\to }{Q_1{P}_1}})$ (alternate interior angles), so $\beta =\delta$. We can deduce of these two equalities that

$$𝜃=\alpha +\beta .$$

(c)

The reflection $r$ with regard to the line $l_1$ sends $P_1$ on $P_2$ and $Q_1$ on $Q_1$, therefore $Q_1{P}_1=Q_1{P}_2$, so the triangle $△Q_1{P}_1{P}_2$ is isosceles, and $$(\widehat{\stackrel{\to }{Q_{1}P},\stackrel{\to }{Q_1{P}_2}})=-(\widehat{\stackrel{\to }{r(Q_1)r(P)},\stackrel{\to }{r(P_1)r(Q_1)}})=-(\widehat{\stackrel{\to }{Q_{1}P},\stackrel{\to }{Q_1{P}_1}}),$$

so the absolute value of the measures of these angles are the same, so

$$\beta =\gamma .$$

(d)

The reflection $s$ of part (a) sends $Q$ on $Q$, and $P_1$ on $Q_1$, therefore $Q{P}_1=Q{Q}_1$. The triangle $△Q{P}_1{Q}_1$ is isosceles, so the corresponding angles are equal: $$-(\widehat{\stackrel{\to }{P_1{Q}_1},\stackrel{\to }{P_{1}Q}})=(\widehat{\stackrel{\to }{Q_1{P}_1},\stackrel{\to }{Q_{1}Q}})=(\widehat{\stackrel{\to }{Q_1{P}_1},\stackrel{\to }{Q_{1}P}})+(\widehat{\stackrel{\to }{Q_{1}P},\stackrel{\to }{Q_{1}Q}}),$$

so

$$\alpha =\beta +\gamma .$$

(e)

From $\delta =\beta$ and from the three equalities obtained in parts (b),(c),(d): $$\begin{array}{llll}\hfill 𝜃& =\alpha +\beta ,& \hfill & \\ \hfill \beta & =\gamma ,& \hfill & \\ \hfill \alpha & =\beta +\gamma ,& \hfill & \end{array}$$

we conclude that $\alpha =2\beta$, $𝜃=3\beta =3\delta$, so $\alpha =2𝜃∕3$, and

$$\delta =𝜃∕3.$$

This means that the measure of the angle formed by the bottom of the square and $\overline{P_1{Q}_1}$ is $𝜃∕3$.