---
title: "Axial Symmetry"
book: "Primary & Middle School Mathematics"
subject: math
language: en
chapter: 42
exercises: 12
source: https://one-course.com/books/math/1/en/chapter/42-axial-symmetry
---

# Chapter 42 — Axial Symmetry

Fold a sheet of paper along a line: every point lands on another point, its mirror image. This folding is *[axial symmetry](#def-g6-symmetry-def)* (or reflection), the first geometric transformation of the book — the second, central symmetry, comes in [Chapter 52](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#ch-g7-central).

## 42.1 Reflections

**Definition 42.1 (Symmetric point).**

Let $d$ be a line. The *symmetric* of a point $M$ across $d$ is the point $M'$ such that $d$ is the [perpendicular bisector](#def-g6-symmetry-bisector) of $[MM']$: the [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) $[MM']$ is [perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to $d$, and $d$ cuts it at its midpoint. A point *on* $d$ is its own symmetric.

![Reflecting across the line d: the axis is perpendicular to (MM') and passes through its midpoint (equal tick marks). A point P on the axis does not move.](https://one-course.com/images/onecourse/chapters/math-1/g6-symmetry/fig-c8d65390f364.svg)

*Reflecting across the line $d$: the axis is [perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to $[MM']$ and passes through its midpoint (equal tick marks). A point $P$ on the axis does not move.*

**Method 42.2 (Constructing a symmetric point).**

To construct the [symmetric](#def-g6-symmetry-def) $M'$ of $M$ across the line $d$:

1. draw the [perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to $d$ through $M$ (set square); call $H$ the point where it crosses $d$ ;
2. measure $MH$ with the compass or ruler;
3. place $M'$ on the same [perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) , on the other side of $d$ , with $HM' = MH$ .

On grid paper it is faster: count the squares from $M$ to the axis, and count the same number on the other side.

**Example 42.3 (On a grid).**

If the axis is a vertical grid line and $M$ is $3$ squares to its left, then $M'$ is $3$ squares to its right, on the same row. For a diagonal axis, the counting works along diagonals — always [perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to the axis.

![A figure and its reflection across the vertical axis: each vertex is reflected square by square, and left and right are swapped.](https://one-course.com/images/onecourse/chapters/math-1/g6-symmetry/fig-1e4ef1df460d.svg)

*A figure and its reflection across the vertical axis: each [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) is reflected square by square, and left and right are swapped.*

**Proposition 42.4 (What a reflection preserves).**

The [symmetric](#def-g6-symmetry-def) of a figure across a line is a figure of the *same shape and same size*: lengths, angles, [perimeters](https://one-course.com/books/math/1/en/chapter/28-measures-and-perimeter#def-g4-measure-perimeter) and [areas](https://one-course.com/books/math/1/en/chapter/28-measures-and-perimeter#def-g4-measure-area) are all preserved. The [symmetric](#def-g6-symmetry-def) of a [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) is a [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects), of a circle a circle of the same [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes), of a line a line.

**Proof.** *Admitted at this level.* ∎

**Remark 42.5.**

One thing is *not* preserved: orientation. A reflected “b” becomes a “d” — mirror writing. If tracing your figure and flipping the tracing paper reproduces the image, the reflection is correct.

## 42.2 Axes of symmetry of a figure

**Definition 42.6 (Axis of symmetry).**

A line $d$ is an *axis of symmetry* of a figure when the reflection across $d$ sends the figure exactly onto itself — folding along $d$ makes the two halves match.

![Axes of symmetry of familiar shapes. Careful with the rectangle: its diagonals are not axes of symmetry (fold along a diagonal — the halves do not match unless it is a square).](https://one-course.com/images/onecourse/chapters/math-1/g6-symmetry/fig-21350edc76df.svg)

*Axes of symmetry of familiar shapes. Careful with the rectangle: its diagonals are *not* axes of symmetry (fold along a diagonal — the halves do not match unless it is a square).*

**Example 42.7.**

Counting axes of symmetry:

- [isosceles triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) : $1$ (through the top [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) and the middle of the base); [equilateral triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) : $3$ ;
- rectangle: $2$ (the two “middle lines”, *not* the diagonals); [rhombus](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-quadrilaterals) : $2$ (its diagonals!); square: $4$ ;
- circle: every line through the center — infinitely many.

**Example 42.8 (Symmetry proves equalities).**

Why are the two base angles of an [isosceles triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) equal? Fold the triangle along its [axis of symmetry](#def-g6-symmetry-axis): the left [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) lands exactly on the right [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half), so the left base angle lands on the right base angle — they must have the same measure ([Proposition 42.4](#prop-g6-symmetry-preserved): reflections preserve angles).

## 42.3 Perpendicular bisector

**Definition 42.9 (Perpendicular bisector).**

The *perpendicular bisector* of a [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) $[AB]$ is the line [perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to $[AB]$ through its midpoint — it is the [axis of symmetry](#def-g6-symmetry-axis) that swaps $A$ and $B$.

**Proposition 42.10 (Equidistance).**

A point $M$ is on the [perpendicular bisector](#def-g6-symmetry-bisector) of $[AB]$ exactly when it is at the same distance from $A$ and from $B$: $MA = MB$.

**Proof.** *Admitted at this level.* ∎

**Method 42.11 (Compass construction).**

To draw the [perpendicular bisector](#def-g6-symmetry-bisector) of $[AB]$ without measuring:

1. open the compass to more than [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) of $AB$ ;
2. draw two arcs centered at $A$ , above and below the [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) ;
3. with the *same* opening, draw two arcs centered at $B$ ;
4. join the two crossing points: this line is the [perpendicular bisector](#def-g6-symmetry-bisector) (each crossing point is equidistant from $A$ and $B$ ).

## 42.4 Exercises

**Exercise 42.1 ★.**

Copy on grid paper: a vertical axis $d$, a point $A$ two squares to its left, a point $B$ five squares to its left and one square higher. Construct their [symmetric](#def-g6-symmetry-def) points $A'$ and $B'$ across $d$.

**Solution of Exercise 42.1.**

$A'$ is two squares to the *right* of $d$, on the same row as $A$; $B'$ is five squares to the right, one square higher — mirror positions.

**Exercise 42.2 ★.**

Draw a line $d$ and a point $M$ at $3$ cm from $d$ (not on grid paper). Construct the [symmetric](#def-g6-symmetry-def) $M'$ of $M$ across $d$ with set square and ruler, following [Method 42.2](#met-g6-symmetry-construct). How far apart are $M$ and $M'$?

**Solution of Exercise 42.2.**

[Perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to $d$ through $M$, foot $H$; then $M'$ on the other side with $HM' = MH = 3$ cm. The distance $MM'$ is $3 + 3 = 6$ cm.

**Exercise 42.3 ★.**

Which capital letters of the alphabet, written in their simplest form, have a vertical [axis of symmetry](#def-g6-symmetry-axis) (like A)? A horizontal one (like B)? Both?

**Solution of Exercise 42.3.**

Vertical axis (like A): A, H, I, M, O, T, U, V, W, X, Y. Horizontal axis (like B): B, C, D, E, H, I, K, O, X. Both: H, I, O, X. (Exact answers depend slightly on how the letters are drawn.)

**Exercise 42.4 ★.**

How many axes of symmetry has: an [equilateral triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles)? a rectangle (that is not a square)? a [rhombus](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-quadrilaterals) (not a square)? a square? a circle? Draw each figure with its axes.

**Solution of Exercise 42.4.**

[Equilateral triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles): $3$ axes. Rectangle: $2$ (the [perpendicular bisectors](#def-g6-symmetry-bisector) of its sides). [Rhombus](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-quadrilaterals): $2$ (its diagonals). Square: $4$. Circle: infinitely many (every line through the center).

**Exercise 42.5 ★.**

True or false? “The diagonals of a rectangle are axes of symmetry of the rectangle.” Explain with a folding argument or a drawing.

**Solution of Exercise 42.5.**

*False.* Folding a (non-square) rectangle along a diagonal does not make the two halves coincide: the two triangles have the same shape but sit differently — try it with a sheet of paper. Only for a square do the diagonals work as axes.

**Exercise 42.6 ★.**

A triangle $ABC$ is reflected across a line $d$ into $A'B'C'$. Given $AB = 4$ cm, $\widehat{BAC} = 70^\circ$ and the [perimeter](https://one-course.com/books/math/1/en/chapter/28-measures-and-perimeter#def-g4-measure-perimeter) of $ABC$ is $12$ cm: give, without any construction, $A'B'$, $\widehat{B'A'C'}$ and the [perimeter](https://one-course.com/books/math/1/en/chapter/28-measures-and-perimeter#def-g4-measure-perimeter) of $A'B'C'$.

**Solution of Exercise 42.6.**

Reflections preserve lengths, angles and [perimeters](https://one-course.com/books/math/1/en/chapter/28-measures-and-perimeter#def-g4-measure-perimeter) ([Proposition 42.4](#prop-g6-symmetry-preserved)): $A'B' = 4$ cm, $\widehat{B'A'C'} = 70^\circ$, [perimeter](https://one-course.com/books/math/1/en/chapter/28-measures-and-perimeter#def-g4-measure-perimeter) of $A'B'C' = 12$ cm.

**Exercise 42.7 ★.**

Draw a [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) $[AB]$ of $7$ cm and construct its [perpendicular bisector](#def-g6-symmetry-bisector) with the compass ([Method 42.11](#met-g6-symmetry-bisector)). Choose any point $M$ on it and check with the ruler that $MA = MB$.

**Solution of Exercise 42.7.**

Compass construction as in [Method 42.11](#met-g6-symmetry-bisector). For any $M$ on the line, the ruler confirms $MA = MB$ ([Proposition 42.10](#prop-g6-symmetry-bisector)).

**Exercise 42.8 ★★.**

Two trees stand at points $A$ and $B$ of a (flat) garden. Where can one plant a fountain so that it is at the same distance from both trees? Describe *all* the possible spots.

**Solution of Exercise 42.8.**

All the points at equal distance from $A$ and $B$ form the [perpendicular bisector](#def-g6-symmetry-bisector) of $[AB]$ ([Proposition 42.10](#prop-g6-symmetry-bisector)). The fountain can go anywhere on that line (within the garden).

**Exercise 42.9 ★★.**

On grid paper, draw the axis $d$ (a diagonal of the grid at $45^\circ$) and a small L-shaped figure on one side. Construct its [symmetric](#def-g6-symmetry-def) figure across $d$. (Reflect each [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def): for a $45^\circ$ axis, a point $2$ squares right of the axis lands $2$ squares above it.)

**Solution of Exercise 42.9.**

Each [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) is reflected [perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to the axis: for the $45^\circ$ axis, a point $k$ squares to the right of the axis lands $k$ squares above it (and vice versa) — horizontal and vertical displacements are exchanged. Reflect the [vertices](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) one by one, then join them in order.

**Exercise 42.10 ★★.**

Draw a circle of center $O$, a line $d$ through $O$, and a point $M$ on the circle. Where is the [symmetric](#def-g6-symmetry-def) of $M$ across $d$? Explain why the reflection sends the circle onto itself.

**Solution of Exercise 42.10.**

The [symmetric](#def-g6-symmetry-def) of $M$ is the second intersection of the circle with the [perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to $d$ through $M$ — it lies on the circle. Reason: the reflection preserves distances and fixes $O$ (which is on $d$), so the image of any point at distance $r$ from $O$ is again at distance $r$ from $O$: the circle maps onto itself.

**Exercise 42.11 ★★.**

Using the equidistance property ([Proposition 42.10](#prop-g6-symmetry-bisector)), explain why the two crossing points of the arcs in [Method 42.11](#met-g6-symmetry-bisector) really do lie on the [perpendicular bisector](#def-g6-symmetry-bisector) of $[AB]$.

**Solution of Exercise 42.11.**

Each crossing point $P$ was drawn with the same compass opening $\rho$ from $A$ and from $B$: so $PA = \rho = PB$. By [Proposition 42.10](#prop-g6-symmetry-bisector), a point equidistant from $A$ and $B$ lies on the [perpendicular bisector](#def-g6-symmetry-bisector) of $[AB]$; two such points determine the line.

**Exercise 42.12 ★★★.**

A billiard ball at point $A$ must bounce off a straight wall $d$ and reach point $B$ (both on the same side of the wall). Reflect $B$ across $d$ into $B'$, and draw the [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) $[AB']$. Explain why the best bouncing point is where $[AB']$ crosses the wall. (Idea: the path $A \to P \to B$ has the same length as $A \to P \to B'$.)

**Solution of Exercise 42.12.**

For any bounce point $P$ on the wall, $PB = PB'$ (reflection across $d$ preserves distances), so the path length $AP + PB = AP + PB'$. The broken line $A \to P \to B'$ is shortest when it is straight, i.e. when $P$ is on the [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) $[AB']$. So the best bounce point is the intersection of $[AB']$ with the wall.

## 42.5 Problem: The fountain, the broken plate, and the walking mirror

**Problem 42.1.**

Weekend problem — the three perpendicular bisectors of a triangle meet at one point: the circle through three points, and what two mirrors do together

In [Exercise 42.8](#exo-g6-symmetry-8) a fountain had to stand at equal distance from *two* trees, and the answer was a whole line of spots. This problem plants a *third* tree — and the whole line collapses to a single, perfectly determined point. That point hides a treasure of geometry: it lets you draw the one circle passing through three given points, rebuild a whole plate from a broken shard, and it belongs to every triangle. The last part performs an experiment with two mirrors whose result announces next year’s geometry.

**Part I — Two trees, then a map of territories.**

1. Draw two points $A$ and $B$ with $AB = 6$ cm and construct the [perpendicular bisector](#def-g6-symmetry-bisector) of $[AB]$ with the compass ( [Method 42.11](#met-g6-symmetry-bisector) ). Pick a point $M$ on it and check with the ruler that $MA = MB$ .
2. [Proposition 42.10](#prop-g6-symmetry-bisector) says two things at once: every point *on* the bisector is equidistant from $A$ and $B$ , and every equidistant point *is on* the bisector. Which of the two answers the fountain question of [Exercise 42.8](#exo-g6-symmetry-8) ? What does the proposition say about a point that is *not* on the bisector?
3. Take a point $M$ on the same side of the bisector as $A$ . The [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) $[MB]$ crosses the bisector at a point $P$ . Compare the path $M \to P \to B$ with the path $M \to P \to A$ , and explain why $M$ is closer to $A$ than to $B$ : the bisector splits the sheet into an “ $A$ -side” and a “ $B$ -side”.
4. Two schools $A$ and $B$ serve a town. Every child walks to the closer school. Draw a small map with $A$ and $B$ , and shade the exact region of the town served by school $A$ . What line forms the border?
5. A third school $C$ opens (place it so the three schools make a triangle). Construct the [perpendicular bisector](#def-g6-symmetry-bisector) of $[AB]$ and the [perpendicular bisector](#def-g6-symmetry-bisector) of $[BC]$ , and call $O$ their crossing point. What two equalities of distances does $O$ satisfy, and why?

**Part II — Three trees, one point, one circle.**

6. Deduce from question 5 that $OA = OC$ — and therefore that $O$ also lies on the [perpendicular bisector](#def-g6-symmetry-bisector) of $[AC]$ , which you never drew. Conclude: *the three [perpendicular bisectors](#def-g6-symmetry-bisector) of the sides of a triangle all pass through one point.*
7. Explain why the circle of center $O$ and [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) $OA$ passes through $B$ and through $C$ as well: it is *the* circle through the three points — the *circumscribed circle* of the triangle $ABC$ .
8. Construct a triangle with sides $6$ cm, $5$ cm and $4$ cm ( [Method 41.2](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#met-g6-shapes-construct) ), then its [circumscribed circle](#pb-g6-symmetry-1) . Why is it enough to draw only *two* [perpendicular bisectors](#def-g6-symmetry-bisector) to find the center?
9. The broken plate: an archaeologist digs up a shard whose only intact part is a piece of the plate’s circular rim. Describe a recipe that recovers the center and the [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) of the original plate — and test it: draw an arc of circle with your compass, “forget” the center, mark three points on the arc, and rebuild it.
10. Could three points ever have *no* circle through them? Suppose $A$ , $B$ , $C$ are aligned. What can be said of the [perpendicular bisectors](#def-g6-symmetry-bisector) of $[AB]$ and of $[BC]$ ( [Proposition 40.3](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#prop-g6-lines-facts) )? Conclude.

**Part III — Axes everywhere, and the walking mirror.**

11. [Example 42.7](#ex-g6-symmetry-shapes) counted $3$ axes for the [equilateral triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) and $4$ for the square. Conjecture the count for a regular [pentagon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) and a regular [hexagon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) , and describe where the axes pass in each case ( [odd number](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) of sides against [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) : the two situations differ).
12. Fold a sheet of paper once and cut a shape through both layers; unfold. Why does the hole *always* have an [axis of symmetry](#def-g6-symmetry-axis) , whatever you cut, and which line is the axis?
13. Fold the sheet twice — the second fold [perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to the first — cut, and unfold. How many axes of symmetry does the hole have at least, and how many copies of your cut do you see? (Paper snowflakes are this idea folded further.)
14. The walking mirror: on grid paper, draw a vertical axis $d_1$ , a small flag $F$ to its left, and a second vertical axis $d_2$ , $4$ squares to the right of $d_1$ . Reflect $F$ across $d_1$ (image $F'$ ), then reflect $F'$ across $d_2$ (image $F''$ ). Compare $F$ and $F''$ : same orientation or mirrored? By how many squares did the figure move? Compare with the distance between the axes.
15. Same experiment with $d_2$ *[perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp)* to $d_1$ : reflect the flag across the vertical axis, then the image across the horizontal one. Describe how $F''$ sits relative to $F$ and to the crossing point of the two axes. (You have just discovered next year’s transformation: [Chapter 52](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#ch-g7-central) .)

**Solution of Problem 42.1.**

**1.** Construction as in [Method 42.11](#met-g6-symmetry-bisector); for any $M$ on the line, the ruler confirms $MA = MB$ ([Proposition 42.10](#prop-g6-symmetry-bisector)).

**2.** The fountain question needs the second direction: *all* the equidistant points are on the bisector, so the possible spots are exactly that line and nothing else. And a point not on the bisector is therefore *not* equidistant: it is strictly closer to one of the two trees.

**3.** The two paths share the leg $[MP]$, and $PB = PA$ since $P$ is on the bisector. So the path $M \to P \to B$ has the same length as the path $M \to P \to A$, namely $MP + PA$. Now $MB$ is the straight route to $B$ — here it *is* the path through $P$, so $MB = MP + PA$; and the straight route to $A$ is shorter than the detour through $P$: $MA < MP + PA = MB$. Every point on $A$’s side of the bisector is closer to $A$.

**4.** The border is the [perpendicular bisector](#def-g6-symmetry-bisector) of $[AB]$: school $A$ serves exactly the [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) of the town on its side of that line (question 3), school $B$ the other [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half), and children living on the line itself may pick either.

**5.** $O$ lies on the bisector of $[AB]$, so $OA = OB$; and on the bisector of $[BC]$, so $OB = OC$ ([Proposition 42.10](#prop-g6-symmetry-bisector)).

**6.** From $OA = OB$ and $OB = OC$: $OA = OC$. By the second direction of [Proposition 42.10](#prop-g6-symmetry-bisector), a point equidistant from $A$ and $C$ lies on the [perpendicular bisector](#def-g6-symmetry-bisector) of $[AC]$: the third bisector passes through $O$ without being asked. The three [perpendicular bisectors](#def-g6-symmetry-bisector) of any triangle are *concurrent*.

**7.** The circle of center $O$ and [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) $OA$ consists of all points at distance $OA$ from $O$ ([Definition 40.5](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-circle)); since $OB = OA$ and $OC = OA$, the points $B$ and $C$ are on it. One center, one [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes), three points served: it is the circle through $A$, $B$ and $C$.

**8.** Two bisectors suffice because the third is automatic (question 6): their crossing point already satisfies all three equalities of distance. The circle is drawn with center $O$ and the compass opened from $O$ to any one of the three [vertices](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def).

**9.** Recipe: mark three points $A$, $B$, $C$ on the surviving arc; draw the [perpendicular bisectors](#def-g6-symmetry-bisector) of the chords $[AB]$ and $[BC]$; their crossing point is the center of the plate, and its distance to $A$ is the [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes). It works because the plate’s rim is a circle through the three marked points — and question 8 finds *the* circle through them. The test arc rebuilds exactly.

**10.** If $A$, $B$, $C$ are aligned, the bisector of $[AB]$ and the bisector of $[BC]$ are both [perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to the same line $(AB) = (BC)$ — so they are [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to each other ([Proposition 40.3](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#prop-g6-lines-facts)) and, being distinct, never cross. No point is equidistant from all three, and no circle passes through three aligned points.

**11.** Regular [pentagon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon): $5$ axes, each through one [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) and the midpoint of the opposite side ([odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) count: every axis is of this one kind). Regular [hexagon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon): $6$ axes — three through opposite [vertices](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) and three through midpoints of opposite sides ([even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) count: two kinds, [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) and [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half)). The pattern: a regular [polygon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) with $n$ sides has $n$ axes.

**12.** The cut goes through both layers at once, so the two layers lose exactly matching pieces; unfolding is the reflection across the fold line, which sends each layer’s edge onto the other’s. The hole is its own mirror image: the fold line is its [axis of symmetry](#def-g6-symmetry-axis) — whatever was cut.

**13.** The unfolded hole has at least $2$ axes of symmetry, the two fold lines, and the cut appears in $4$ copies (each unfolding doubles the picture: $1 \to 2 \to 4$). Folding more times in the snowflake style multiplies the copies and the axes further.

**14.** $F''$ has the *same* orientation as $F$ (two mirror flips cancel the mirroring), the same size and the same height — it is simply $F$ slid to the right. The slide measures $8$ squares: exactly *twice* the distance between the two axes. Two [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) mirrors together do not mirror at all: they *translate*.

**15.** $F''$ is again the same size, but now it appears *upside down*, turned by [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) a turn about the crossing point of the two axes: each point of $F''$ is diametrically opposite the corresponding point of $F$ through that center. Two [perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) mirrors compose into the *half-turn* — the central symmetry of [Chapter 52](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#ch-g7-central).