Line and Rotational Symmetry

A line of symmetry folds a figure exactly onto itself, and rotational symmetry turns it onto itself. Find the lines of symmetry and the order of rotational symmetry of letters, simple designs and the three kinds of triangle.

Before you start

A triangle is scalene if no two of its sides are equal, isosceles if at least two of its sides are equal, and equilateral if all three of its sides are equal. Equal sides are marked with the same number of small ticks. An equilateral triangle is isosceles too (Triangles and Their Angles).
In an isosceles triangle the two equal sides meet at the apex; the side opposite the apex is the base, and the two base angles are equal (base ∠s of isos. △). If two angles of a triangle are equal, the sides opposite them are equal (isos. △). Each angle of an equilateral triangle is 60°, and a triangle with three equal angles is equilateral (equilateral △) (Isosceles and Equilateral Triangles).
The angles of a triangle add up to 180° (∠ sum of △), and adjacent angles on a straight line add up to 180° (adj. ∠s on a str. line).

Line Symmetry

Fold a sheet of paper in half, cut out a shape across the fold and open it out: the two halves of the shape match exactly, one on each side of the fold. The fold line is a line of symmetry of the shape.

Symmetry describes how a figure can be folded or turned so that it fits exactly onto itself. This lesson teaches it on letters, simple designs and triangles; it is used again for every shape in this chapter.

Line of symmetry

A line of symmetry divides a figure into two parts that are mirror images of each other.

When the figure is folded along this line, one part fits exactly onto the other.

On this page a line of symmetry is drawn dashed, in blue.

Two parts of the same shape and size are not enough: they must be mirror images, so that the fold puts each part exactly onto the other. To test a line, imagine folding along it, or stand a mirror on it and see whether the half you can see, with its reflection, makes the whole figure.

Example 1: Lines of symmetry of letters

How many lines of symmetry has each of the block letters T, E and N? Draw them.

Solution

T: 1 line

E: 1 line

N: none

T. Fold along the upright line through the middle of the stem: the left half of the top bar fits onto the right half, and the two halves of the stem fit onto each other. No other fold works: folded across, the top bar would land at the foot of the stem, where there is none.

E. Fold along the line across the middle of the middle arm: the top arm fits onto the bottom arm. An upright fold does not work, because all three arms point to the right and their mirror images would point to the left.

N. The upright line through its middle cuts N into two parts of the same shape and size, but they are not mirror images. Folded along that line, the left post fits onto the right post, but the slanting stroke, which runs down from the top left to the bottom right, lands where a stroke from the top right to the bottom left would be, and there is none. Every other fold fails in the same way.

T has 1 line of symmetry, E has 1 and N has none.

Lines of Symmetry of a Triangle

In the lesson Isosceles and Equilateral Triangles, an isosceles triangle $ABC$ with $AB = AC$ was folded through $A$ so that $AB$ lay along $AC$ . Because $AB = AC$ , the point $B$ landed exactly on $C$ and the two halves fitted: the fold line is a line of symmetry of the triangle.

Isosceles (not equilateral): 1 line

Equilateral: 3 lines

Scalene: no line of symmetry

An isosceles triangle that is not equilateral has exactly one line of symmetry: the line through the apex and the midpoint of the base.

An equilateral triangle has three lines of symmetry. Any of its vertices can be taken as the apex, so there is a line from each vertex to the midpoint of the opposite side.

A scalene triangle (no two sides equal) has no line of symmetry. However you fold it, no side fits onto another.

In every case, a line of symmetry of a triangle runs from a vertex to the midpoint of the opposite side, and the fold puts one of the sides at that vertex onto the other. So each line of symmetry of a triangle shows two equal sides.

What the fold tells you about an isosceles triangle

In $\triangle ABC$ with $AB = AC$ , let the line of symmetry through $A$ meet $BC$ at $M$ . Folding along $AM$ puts each half of the triangle exactly onto the other, so the line of symmetry:

(1) cuts the apex angle into two equal angles: the fold puts $\angle BAM$ onto $\angle CAM$ ;

(2) cuts the base into two equal parts: the fold puts $B$ onto $C$ , so $MB$ fits exactly onto $MC$ ;

(3) meets the base at right angles: the fold puts $\angle AMB$ onto $\angle AMC$ , so they are equal, and they lie together on the straight line $BC$ , so each is $180^{\circ} \div 2 = 90^{\circ}$ .

Example 2: Angles at the line of symmetry

In $\triangle PQR$ , $PQ = PR$ and $\angle QPR = 40^{\circ}$ . The dashed line $PM$ is the line of symmetry of the triangle, and $M$ lies on $QR$ .

Find (a) $\angle QPM$ , (b) $\angle PMQ$ and (c) $\angle PQM$ .

Solution

(a) Folding along $PM$ puts $PQ$ onto $PR$ , so $\angle QPM$ fits exactly onto $\angle RPM$ : they are equal. Together they make $\angle QPR$ , so

$\angle QPM = 40^{\circ} \div 2 = 20^{\circ}$ ( $PM$ is a line of symmetry)

(b) The fold also puts $\angle PMQ$ onto $\angle PMR$ , so they are equal. They lie together on the straight line $QR$ , so

$\angle PMQ = 180^{\circ} \div 2 = 90^{\circ}$ ( $PM$ is a line of symmetry; adj. ∠s on a str. line)

(c) $\angle PQM = \angle PRM$ (base ∠s of isos. △), so $\angle PQM = (180^{\circ} - 40^{\circ}) \div 2 = 70^{\circ}$ (∠ sum of △).

Check in $\triangle PQM$ : $20^{\circ} + 90^{\circ} + 70^{\circ} = 180^{\circ}$ .

(a) $\angle QPM = 20^{\circ}$ (b) $\angle PMQ = 90^{\circ}$ (c) $\angle PQM = 70^{\circ}$ .

Example 3: Can a triangle have exactly two lines of symmetry?

Priya says she has drawn a triangle with exactly two lines of symmetry. Explain why she must be wrong.

Solution

Two different lines of symmetry cannot both run from the same vertex: at vertex $A$ , the only fold that puts $AB$ onto $AC$ is the one that cuts $\angle BAC$ into two equal angles, and there is only one such line. So Priya's two lines run from two different vertices, say $A$ and $B$ .

The line from $A$ puts $AB$ onto $AC$ , so $AB = AC$ . The line from $B$ puts $BA$ onto $BC$ , so $BA = BC$ . So $AB = AC = BC$ , and the triangle is equilateral.

But an equilateral triangle has a third line of symmetry, from $C$ . So no triangle has exactly two.

A triangle with two lines of symmetry is equilateral, so it has three. A triangle has 0, 1 or 3 lines of symmetry, never exactly 2.

Rotational Symmetry

Some figures fit onto themselves when they are turned. Stick a pin through the middle $O$ of the letter N and turn it through half a complete turn (180°): it fits exactly onto the place where it started. Only the corner $P$ , marked with the amber line $OP$ , shows that it has moved.

Start

After a half-turn (180°)

Now turn an equilateral triangle anticlockwise about its centre $O$ , the point where its three lines of symmetry meet. After a third of a complete turn (120°), it fits exactly onto the place where it started (shaded): only the letters, and the amber line $OA$ that turns with it, show that it has moved. It fits again after 240°, and again after a complete turn (360°), when every vertex is back where it began.

Start

After 120° anticlockwise

After 240° anticlockwise

After a complete turn (360°)

Turned clockwise instead, the triangle fits after the same angles (120°, 240° and 360°); only the places the letters go to are different. In this lesson every turn is anticlockwise unless it says otherwise.

Rotational symmetry

A figure has rotational symmetry if it fits exactly onto itself when it is turned about a fixed point, its centre, through an angle less than one complete turn.

The order of rotational symmetry is the number of times the figure fits onto itself during one complete turn, counting the complete turn itself.

The letter N fits onto itself after 180° and after 360°: twice in one complete turn. So it has rotational symmetry of order 2.

The equilateral triangle fits onto itself after 120°, 240° and 360°: three times in one complete turn. So it has rotational symmetry of order 3.

Every figure fits onto itself after a complete turn, so the order is never less than 1. Order 1 means no rotational symmetry: the figure fits onto itself only when it is back where it started. The letter T has no rotational symmetry: turned through 180°, its top bar is at the bottom. In a table we write "order 1"; in a sentence we say the figure "has no rotational symmetry".

The smallest turn

If the smallest turn that fits a figure onto itself is $a^{\circ}$ , it fits again after $2a^{\circ}$ , $3a^{\circ}$ , … (turn it by $a^{\circ}$ again each time), up to the complete turn of 360°.

So the order of rotational symmetry $= 360^{\circ} \div$ the smallest turn, and the smallest turn $= 360^{\circ} \div$ the order.

For the equilateral triangle: $360^{\circ} \div 120^{\circ} = 3$ . For the letter N: $360^{\circ} \div 180^{\circ} = 2$ .

Example 4: A toy windmill

Ali's toy windmill has four blades, all the same, fixed at its centre $O$ .

(a) Find its order of rotational symmetry and the smallest turn that fits it onto itself.

(b) How many lines of symmetry has it?

Solution

(a) A quarter-turn (90°) about $O$ carries each blade onto the next one:

It fits onto itself after 90°, 180°, 270° and 360°: four times in one complete turn, so the order is 4. The smallest turn is $360^{\circ} \div 4 = 90^{\circ}$ .

(b) Every blade leans the same way round. The mirror image of a blade leans the other way round, so whatever line you fold along, no blade lands on a blade.

(a) Order 4; the smallest turn is 90°. (b) It has no line of symmetry.

So a figure can have rotational symmetry and no line of symmetry (the windmill, the letter N), or a line of symmetry and no rotational symmetry (the letters T and E).

Rotational Symmetry of a Triangle

Why a turn that fits a triangle moves every vertex

When a triangle fits onto itself, each vertex lands on a vertex: a corner can only cover a corner.

A triangle turned about one of its own corners, through less than a complete turn, swings off its place. So a turn short of a complete turn that fits a triangle onto itself is about a point inside it, and moves every vertex onto another vertex.

So the angle at each vertex must equal the angle at the vertex it lands on.

In the equilateral triangle all three angles are 60°, so the 120° turn can carry each vertex onto the next.

An isosceles triangle that is not equilateral has no rotational symmetry. A turn that fitted it would move the apex onto a base vertex, but the apex angle is different from the base angles (if it were equal to them, all three angles would be equal and the triangle would be equilateral). The figure shows it after a turn of 120° about a point $O$ inside it: it no longer covers its starting place. Turned about any other point, or through any other angle short of a complete turn, it fails in the same way.

Start

After 120° anticlockwise: it does not fit

A scalene triangle has no rotational symmetry either: its three angles are all different, so no vertex can go onto another, and a turn short of a complete turn must move them.

The symmetry of the three kinds of triangle:

Triangle	Lines of symmetry	Order of rotational symmetry
Scalene	0	1
Isosceles (not equilateral)	1	1
Equilateral	3	3

The symmetry of the special quadrilaterals and of the regular polygons is found in the same way, and comes in later lessons of this chapter (Properties of Special Quadrilaterals; Polygons and Regular Polygons).

Common Mistakes

Calling any line that cuts a figure into two equal parts a line of symmetry. The two parts must be mirror images. The upright line through the letter N cuts it into two parts of the same shape, but folded along it they do not fit. In the same way, a line from a vertex of a scalene triangle to the midpoint of the opposite side is not a line of symmetry.
Saying "order 0". Every figure fits onto itself after a complete turn, so the order of rotational symmetry is at least 1. A figure with no rotational symmetry has order 1.
Counting the start and the complete turn as two. The equilateral triangle is in its starting place at 0° and again at 360°: that is one fit, not two. Count the fits after 120°, 240° and 360°: order 3, not 4.
Thinking a figure with no line of symmetry has no rotational symmetry. The letter N and the toy windmill have no line of symmetry, but they have rotational symmetry of order 2 and order 4.
Forgetting "not equilateral". An equilateral triangle is isosceles too, and it has 3 lines of symmetry. "1 line of symmetry, order 1" is true of an isosceles triangle that is not equilateral.

Practice

Try each question, then open its solution to check your working. The figures are drawn to scale.

Practice 1: Lines of symmetry of letters

How many lines of symmetry has each of the block letters V, F and H?

ShowHide solution

V: 1 line

F: none

H: 2 lines

V. Folded along the upright line through its middle, its left stroke fits onto its right stroke. Folded across, the point at the bottom would land at the top, where the V is open: no other line works.

F. Both arms point to the right, and the arms are at the top, not the bottom: neither an upright fold nor a fold across puts the F onto itself.

H. The upright line puts the left post onto the right post, and the line across the middle of the crossbar puts the top half onto the bottom half. The H is taller than it is wide, so a fold along a slanting line would put a post where there is none.

V has 1 line of symmetry, F has none and H has 2.

Practice 2: Rotational symmetry of letters

State the order of rotational symmetry of each of the block letters H, Z and V. For each letter that has rotational symmetry, state the smallest turn that fits it onto itself.

ShowHide solution

H. Turned through 180° about its middle, the left post goes onto the right post and the right post onto the left, and the crossbar onto itself: it fits. It fits again after 360°. A turn of 90° would lay the posts on their sides, so it fits only twice in a complete turn: order 2, smallest turn $360^{\circ} \div 2 = 180^{\circ}$ .

Z. Turned through 180° about its middle, the top bar goes onto the bottom bar and the slanting stroke onto itself: it fits. It fits again after 360°, and after no other turn: order 2, smallest turn 180°.

V. Turned through 180°, its point would be at the top: it fits onto itself only after a complete turn. It has no rotational symmetry: order 1.

H: order 2, smallest turn 180°. Z: order 2, smallest turn 180°. V: order 1 (no rotational symmetry).

Practice 3: Triangles from their angles

For each triangle, say whether it is scalene, isosceles (not equilateral) or equilateral, and state its number of lines of symmetry and its order of rotational symmetry.

(a)

(b)

(c)

ShowHide solution

(a) $\angle PQR = \angle PRQ = 65^{\circ}$ , so the sides opposite them are equal: $PR = PQ$ (isos. △). Its angles are not all 60°, so it is not equilateral. An isosceles triangle that is not equilateral has 1 line of symmetry (through $P$ and the midpoint of $QR$ ) and no rotational symmetry: order 1.

(b) Its three angles are all different. If two of its sides were equal, the angles opposite them would be equal (base ∠s of isos. △); so no two sides are equal, and it is scalene: no line of symmetry, and no rotational symmetry: order 1.

(c) $\angle YXZ = 180^{\circ} - 60^{\circ} - 60^{\circ} = 60^{\circ}$ (∠ sum of △). All three angles are 60°, so it is equilateral (equilateral △): 3 lines of symmetry and rotational symmetry of order 3.

(a) Isosceles (not equilateral): 1 line, order 1. (b) Scalene: no line, order 1. (c) Equilateral: 3 lines, order 3.

Practice 4: A road sign

A road sign that warns drivers of danger ahead is in the shape of an equilateral triangle. Think only of the triangle, not of the picture on it.

(a) What is the size of each angle of the triangle?

(b) How many lines of symmetry does it have?

(c) What is its order of rotational symmetry, and what is the smallest turn about its centre that makes it fit onto itself?

ShowHide solution

(a) Each angle of an equilateral triangle is 60° (equilateral △).

(b) It has 3 lines of symmetry, each from a vertex to the midpoint of the opposite side:

(c) Turned about its centre, it fits onto itself after 120°, 240° and 360°: three times in one complete turn, so the order is 3. The smallest turn is $360^{\circ} \div 3 = 120^{\circ}$ .

(a) 60° (b) 3 lines of symmetry (c) order 3; the smallest turn is 120°.

Practice 5: Set squares

Mei Ling has two set squares. The angles of one are 45°, 45° and 90°; the angles of the other are 30°, 60° and 90°.

Set square 1

Set square 2

For each set square, (a) say whether it is scalene, isosceles or equilateral, (b) state its number of lines of symmetry, and (c) state its order of rotational symmetry.

ShowHide solution

Set square 1. (a) Two of its angles are equal (45° and 45°), so the sides opposite them are equal (isos. △): it is isosceles (an isosceles right-angled triangle). Its angles are not all 60°, so it is not equilateral.

(b) An isosceles triangle that is not equilateral has 1 line of symmetry: through the right-angled corner (the apex) and the midpoint of the side opposite the right angle (the base).

Set square 2. (a) Its three angles are all different. If two of its sides were equal, the angles opposite them would be equal (base ∠s of isos. △); so no two sides are equal, and it is scalene.

(b) A scalene triangle has no line of symmetry. (c) It has no rotational symmetry: order 1.

Set square 1: isosceles, 1 line of symmetry, order 1. Set square 2: scalene, no line of symmetry, order 1.

Practice 6: Turns that fit

(a) The blades of a ceiling fan, all the same, fit onto themselves when they are turned through 72° about the centre of the fan, and through no smaller turn. What is the order of rotational symmetry of the blades, and how many blades has the fan?

(b) A design has rotational symmetry of order 6. List every turn, up to one complete turn, after which it fits onto itself.

(c) Wei Ming says that his design fits onto itself after a turn of 50°, and after no smaller turn. Explain why he must be wrong.

ShowHide solution

(a) The blades fit after 72°, 144°, 216°, 288° and 360°: the order is $360^{\circ} \div 72^{\circ} = 5$ . Each turn of 72° carries every blade onto the next one, and five such turns make a complete turn, so the fan has 5 blades.

(b) The smallest turn is $360^{\circ} \div 6 = 60^{\circ}$ , so it fits after $60^{\circ}$ , $120^{\circ}$ , $180^{\circ}$ , $240^{\circ}$ , $300^{\circ}$ and $360^{\circ}$ .

(c) If it fits after 50°, it fits after every further turn of 50°: after $7 \times 50^{\circ} = 350^{\circ}$ and after $8 \times 50^{\circ} = 400^{\circ}$ . A turn of 400° is a complete turn and then 40° more, and the complete turn brings the design back to its start, so it would also fit after a turn of 40°. That is smaller than 50°, so 50° cannot be the smallest turn. (The smallest turn must go into 360° a whole number of times, and $360 \div 50 = 7.2$ .)

(a) Order 5; the fan has 5 blades. (b) 60°, 120°, 180°, 240°, 300° and 360°. (c) It would also fit after 40°, so 50° is not the smallest turn: the smallest turn must go into 360° exactly.

Practice 7: A tent

The front of Siti's tent is a triangle $ABC$ . Its two sloping sides $AB$ and $AC$ are equal, and the side $BC$ lies on the flat ground. Each sloping side makes an angle of 62° with the ground. A pole $AM$ joins the top $A$ to the point $M$ on the ground, along the line of symmetry of the triangle (the pole is drawn in amber).

Find (a) $\angle BAC$ , the angle at the top of the tent, (b) $\angle BAM$ , the angle between the pole and a sloping side, and (c) $\angle AMB$ , the angle between the pole and the ground.

ShowHide solution

(a) $\angle BAC = 180^{\circ} - 62^{\circ} - 62^{\circ} = 56^{\circ}$ (∠ sum of △)

(b) The pole lies along the line of symmetry, which cuts the apex angle into two equal angles:

$\angle BAM = 56^{\circ} \div 2 = 28^{\circ}$ ( $AM$ is a line of symmetry)

(c) The fold along $AM$ puts $\angle AMB$ onto $\angle AMC$ , so they are equal, and they lie together on the straight line $BC$ :

$\angle AMB = 180^{\circ} \div 2 = 90^{\circ}$ ( $AM$ is a line of symmetry; adj. ∠s on a str. line)

Check in $\triangle ABM$ : $62^{\circ} + 28^{\circ} + 90^{\circ} = 180^{\circ}$ .

(a) $\angle BAC = 56^{\circ}$ (b) $\angle BAM = 28^{\circ}$ (c) $\angle AMB = 90^{\circ}$ : the pole stands upright.

Practice 8: The two halves of the apex angle

In $\triangle ABC$ , $AB = AC$ . The dashed line $AM$ is the line of symmetry of the triangle, with $M$ on $BC$ . $\angle BAM = (2x + 10)^{\circ}$ and $\angle CAM = (3x - 5)^{\circ}$ .

Find (a) $x$ , (b) $\angle BAC$ , (c) $\angle ABC$ and (d) $\angle AMC$ .

ShowHide solution

(a) The fold along $AM$ puts $\angle BAM$ onto $\angle CAM$ , so they are equal ( $AM$ is a line of symmetry):

$3x - 5 = 2x + 10$

Subtract $2x$ from both sides: $x - 5 = 10$

Add 5 to both sides: $x = 15$

(b) $\angle BAM = (2 \times 15 + 10)^{\circ} = 40^{\circ}$ and $\angle CAM = (3 \times 15 - 5)^{\circ} = 40^{\circ}$ , so $\angle BAC = 40^{\circ} + 40^{\circ} = 80^{\circ}$ .

(c) $\angle ABC = \angle ACB$ (base ∠s of isos. △), so $\angle ABC = (180^{\circ} - 80^{\circ}) \div 2 = 50^{\circ}$ (∠ sum of △).

(d) $\angle AMC = 180^{\circ} \div 2 = 90^{\circ}$ ( $AM$ is a line of symmetry; adj. ∠s on a str. line)

Check in $\triangle AMC$ : $40^{\circ} + 50^{\circ} + 90^{\circ} = 180^{\circ}$ .

(a) $x = 15$ (b) $\angle BAC = 80^{\circ}$ (c) $\angle ABC = 50^{\circ}$ (d) $\angle AMC = 90^{\circ}$ .

Practice 9: A window grille

The middle of a window grille in an HDB flat has the steel design below: a frame in the shape of an equilateral triangle, and three bent bars (drawn in amber), all the same, joined at the centre of the frame.

(a) How many lines of symmetry has the frame on its own, and what is its order of rotational symmetry?

(b) How many lines of symmetry has the whole design, and what is its order of rotational symmetry? What is the smallest turn that fits it onto itself?

ShowHide solution

(a) The frame on its own is an equilateral triangle: 3 lines of symmetry and rotational symmetry of order 3.

(b) A turn of 120° about the centre carries the frame onto itself and each bent bar onto the next one, so the design fits after 120°, 240° and 360°: order 3, and the smallest turn is $360^{\circ} \div 3 = 120^{\circ}$ .

Each bar bends the same way round. A fold along any of the frame's three lines of symmetry would carry each bar onto its mirror image, which bends the other way round, where there is no bar. A fold that fitted the whole design onto itself would have to fit the frame onto itself, so it could only be one of these three lines; so the design has no line of symmetry.

(a) 3 lines of symmetry, order 3. (b) No line of symmetry; order 3; the smallest turn is 120°.

Practice 10: Wall tiles

Mr Tan's bathroom wall has tiles in the shape of equilateral triangles. On tile 1 an amber stripe joins $M$ , the midpoint of $AB$ , to $N$ , the midpoint of $BC$ . On tile 2 the triangle joining the midpoints of the three sides is coloured amber.

Tile 1

Tile 2

For each tile with its pattern, state the number of lines of symmetry and the order of rotational symmetry.

ShowHide solution

Tile 1. A fold that fits the tile with its pattern onto itself must fit the triangle $ABC$ onto itself, so only the three lines of symmetry of the equilateral triangle can work. Test each of them.

The line through $B$ and $K$ , the midpoint of $CA$ , puts $A$ onto $C$ , so it puts $AB$ onto $CB$ and the midpoint $M$ of $AB$ onto the midpoint $N$ of $CB$ . The stripe $MN$ folds exactly onto itself: this is a line of symmetry.

The line through $A$ puts $B$ onto $C$ , so it carries the stripe to the line joining the midpoints of $AC$ and $BC$ , where there is no stripe; the line through $C$ fails in the same way.

A turn that fits the tile onto itself must fit the triangle $ABC$ onto itself, so it can only be a turn of 120° or 240° about the centre of the triangle. Each of these carries the stripe to join the midpoints of a different pair of sides, where there is none, so tile 1 has no rotational symmetry.

Tile 2. Each of the tile's three lines of symmetry puts the midpoints of the sides onto midpoints of sides, so it folds the amber triangle onto itself. A turn of 120° about the centre carries each side onto the next, and its midpoint onto the next midpoint, so the amber triangle fits onto itself after 120°, 240° and 360°.

Tile 1: 1 line

Tile 2: 3 lines

Tile 1: 1 line of symmetry, order 1 (no rotational symmetry). Tile 2: 3 lines of symmetry, order 3.

Practice 11: Is it equilateral?

In $\triangle ABC$ , $AB = AC$ . The dashed line $AM$ is the line of symmetry of the triangle, with $M$ on $BC$ . $\angle BAM = (x + 10)^{\circ}$ and $\angle ABM = (2x + 20)^{\circ}$ .

(a) Find $x$ .

(b) Show that $\triangle ABC$ is equilateral.

ShowHide solution

(a) The line of symmetry meets the base at right angles: $\angle AMB = 90^{\circ}$ ( $AM$ is a line of symmetry; adj. ∠s on a str. line).

In $\triangle ABM$ : $(x + 10) + (2x + 20) + 90 = 180$ (∠ sum of △)

Collect like terms: $3x + 120 = 180$

Subtract 120 from both sides: $3x = 60$

Divide both sides by 3: $x = 20$

(b) $\angle BAM = (20 + 10)^{\circ} = 30^{\circ}$ , so $\angle BAC = 2 \times 30^{\circ} = 60^{\circ}$ ( $AM$ is a line of symmetry). $\angle ABC = (2 \times 20 + 20)^{\circ} = 60^{\circ}$ , and $\angle ACB = \angle ABC = 60^{\circ}$ (base ∠s of isos. △).

All three angles are 60°, so $\triangle ABC$ is equilateral (equilateral △).

Check in $\triangle ABM$ : $30^{\circ} + 60^{\circ} + 90^{\circ} = 180^{\circ}$ .

(a) $x = 20$ (b) each angle of $\triangle ABC$ is 60°, so it is equilateral (c) 3 lines of symmetry, order 3.

Practice 12: A card triangle on a pin

Ali cuts a triangle $ABC$ out of card and pins it to a board through a point $O$ inside it. He turns the card anticlockwise about $O$ . After a turn of 120°, the card fits exactly onto the place where it started.

(a) After which other turns, up to one complete turn, does the card fit onto its starting place? State its order of rotational symmetry.

(b) What kind of triangle is $ABC$ ? Give a reason.

ShowHide solution

(a) Turning the card through another 120° fits it again, after 240° in all, and another 120° brings it back to its start after a complete turn (360°). It fits after 120°, 240° and 360°: three times in one complete turn, so its order of rotational symmetry is 3 ( $360^{\circ} \div 120^{\circ} = 3$ ).

(b) Only $O$ stays still, so the turn of 120° moves every corner onto the place of a different corner. Say $A$ moves to where $B$ was. Then $B$ cannot move to where $A$ was, or the only place left for $C$ would be its own; so $B$ moves to where $C$ was, and $C$ to where $A$ was.

A corner can only cover a corner of the same size, so $\angle BAC = \angle ABC$ ( $A$ covers the place of $B$ ) and $\angle ABC = \angle ACB$ ( $B$ covers the place of $C$ ).

All three angles are equal, so $\triangle ABC$ is equilateral (equilateral △).

Check: $60^{\circ} + 60^{\circ} + 60^{\circ} = 180^{\circ}$ .

(a) After 240° and 360°; order 3. (b) Equilateral: the turn carries each corner onto another corner of the same size, so all three angles are equal. (c) Each angle is 60°; 3 lines of symmetry.

Key Points

A line of symmetry divides a figure into two parts that are mirror images; folded along it, one part fits exactly onto the other.
A figure has rotational symmetry if it fits onto itself when turned about its centre through less than a complete turn.
The order of rotational symmetry is the number of times a figure fits onto itself in one complete turn, counting the complete turn; order 1 means no rotational symmetry.
The smallest turn $= 360^{\circ} \div$ the order; the figure fits after every multiple of the smallest turn, up to 360°.
A figure can have rotational symmetry and no line of symmetry (the letter N), or a line of symmetry and no rotational symmetry (the letter T).
A line of symmetry of a triangle runs from a vertex to the midpoint of the opposite side. In an isosceles triangle it cuts the apex angle in half, cuts the base in half and meets the base at right angles.
Scalene: no line of symmetry, order 1. Isosceles (not equilateral): 1 line, order 1. Equilateral: 3 lines, order 3. No triangle has exactly 2 lines.