New Century Maths 10: Stage 5.2/5.3 - Chapter 5

Wordbank

Zdeduction The formal process of proving something logically and reaching a conclusion by reasoning from a general principle to a specific result.

Zparallelogram A quadrilateral with both pairs of opposite sides parallel.

Zrectangle A parallelogram with one angle a right angle.

deductive geometry

deductive geometry

Think!

A square is a rhombus, but a rhombus is not necessarily a square. Why not? Can a square be considered as a rectangle?

S PAC E A N D G E O M E T RY

In this chapter you will:

Zrecognise the types of triangles and quadrilaterals and the properties of their sides, angles and diagonals Zapply simple deductive reasoning to solving

numerical and non-numerical problems

Zconstruct and write geometrical arguments to prove a geometrical result, giving reasons at each step of the argument

Zapply angle sum results to find unknown angles in convex polygons

Zwrite formal proofs of the congruence of triangles, preserving the matching order of the vertices Zprove properties of special triangles and

quadrilaterals from the formal definitions of the shapes

Zprove and apply theorems related to triangles and quadrilaterals

Zprove and apply tests for quadrilaterals

Zcalculate unknown sides in a pair of similar triangles

Zdetermine what information is needed to establish that two triangles are similar

Zwrite formal proofs of the similarity of triangles, preserving the matching order of the vertices, identifyng the similarity factor when appropriate Zsolve Euclidian geometry problems

Zprove and apply further theorems using similarity

Zhence Following from the previous result, another word for ‘therefore’.

Zdeductive geometry When geometrical results and

problems are proved or solved by a step-by-step process or argument. Reasons are given at each step of the argument.

test for quadrilaterals A property of a quadrilateral

Zsimilarity test for triangles One of four tests (SSS, SAS, AA, RHS) that prove that two triangles are similar.

Zconverse The reverse of a geometrical rule or theorem, written in a back-to-front way. For example, the converse of ‘If lines are parallel, corresponding angles are equal’ is ‘If corresponding

1 Find the values of the pronumerals in the following diagrams. Give reasons for your answers.

a b c

d e f

2 Find the value of each pronumeral, giving reasons for each step.

a b c

d e f

3 Find the values of the pronumerals in the following diagrams. Give reasons for your answers.

a b c d e f m° 63° x° 47° 82° y° 78° 135° k° 42° w° 38° 127° 4a° 5a° 37° 152° 127° d° 88° 4m° 66° 2h° a° 72° 104° m° k° 81° w° p° h° 71° 63° m° 35° 124° 3h° 5h° 2h° 2y° 137° 83° 3y° 3k° 105° a° 163° 27° 34°

Start up

Worksheet 5-01 Brainstarters 5 Skillsheet 5-01 Angles and parallel lines

Deduction in geometry

Many results in geometry can be shown or demonstrated by construction and measurement. For example, we can draw a triangle and measure the angles to show or demonstrate that the angle sum of a triangle is 180°. However this does not prove that the angle sum of any triangle is 180°. To prove this and other geometrical results we use a process called deduction, in which a specific result is proved by reasoning logically from a general principle or known fact.

When setting out proofs, reasons must be given for each statement or each step of the argument. The reasons are usually in brackets after the statements and should state the names of the triangles, angles, or lines referred to.

Important results which have been proved and are then used as a rule or general principle are called theorems.

4 Which of these pairs of lines are parallel? Give reasons.

a b c 67° 69° 132° 48° 78° 103°

Euclid and Euclidian geometry

Euclid was a Greek mathematician who lived in Alexandria during the third century BC. Euclid’s most famous written work was the thirteen-volume Elements, which he probably wrote before the age of 25. This work presented a systematic treatment of all known geometry, and actually covered most of the mathematics known at the time. There are 465 different propositions in Elements (Pythagoras’ theorem being number 47 in the first volume). Geometry using the methods from Euclid’s Elements is called Euclidian geometry and the results and proofs that will be considered in the chapter are close to those of Euclid. Other geometries were developed by Gauss, Lobachevsky, Rieman and others in the early nineteenth century. Select one of these geometries and briefly describe how it is different from Euclidean geometry.

Just for the record

Example 1

Find the values of the pronumerals in the following diagrams:

a b k° 55° d° 46° A _{B C} D E

Properties of triangles and quadrilaterals

Triangles

Definitions of special triangles

Properties

The properties of the various types of triangles can be summarised as follows. Scalene

– no equal sides – no equal angles – no symmetry

Isosceles

– two equal sides – two equal angles – one axis of symmetry – no rotational symmetry

Equilateral

– three equal sides

– three equal angles (60° each) – three axes of symmetry – has rotational symmetry

Solution

a k + 55 + 55 = 180 (angle sum of isosceles triangle)

k + 110 = 180 k = 70

b ∠EDC = ∠ECD = (180 − 46) ÷ 2 (angle sum of isosceles ∆ECD)

∠EDC = 67°

∠EBC = 67° (opposite angles of a parallelogram are equal)

d + 67 = 180 (angle sum of a straight line)

∴ d = 113

Skillsheet 5-02

Two-dimensional shapes

• An acute-angled triangle is a triangle with all angles acute. • An obtuse-angled triangle is a triangle with an obtuse angle. • A right-angled triangle is a triangle with a right angle.

• A scalene triangle is a triangle with no two sides equal in length. • An isosceles triangle is a triangle with two sides equal in length. • An equilateral triangle is a triangle with all sides equal in length.

Quadrilaterals

A quadrilateral is a four-sided plane figure. A quadrilateral may be either convex or non-convex (concave).

A convex quadrilateral has no interior angles greater than 180°.

A convex quadrilateral A non-convex quadrilateral

(all angles are less than 180°) (one interior angle is greater than 180°) The diagonals of a convex quadrilateral lie inside the quadrilateral.

Diagonals AC and BD lie inside the The diagonal MP does not lie inside the convex quadrilateral ABCD. non-convex quadrilateral LMNP.

In the work that follows, convex quadrilaterals will be simply referred to as ‘quadrilaterals’.

Definitions of special quadrilaterals

Worksheet 5-02 Naming quadrilaterals Worksheet 5-03 Investigating the geometrical constructions Skillsheet 5-03 Geometrical constructions A B C D L M N P

A trapezium is a quadrilateral with at least one pair of opposite sides parallel. A parallelogram is a quadrilateral with both pairs of opposite sides parallel. A rectangle is a parallelogram with one angle a right angle.

A square is a rectangle with two adjacent sides equal in length. A rhombus is a parallelogram with two adjacent sides equal in length. A kite is a convex quadrilateral with two pairs of equal adjacent sides.

Summary

The properties of the various types of quadrilaterals can be summarised as follows: Trapezium – one pair of opposite sides parallel

Parallelogram – opposite sides parallel – opposite sides equal – opposite angles equal – diagonals bisect each other – has rotational symmetry Rectangle – opposite sides parallel

– opposite sides equal – all angles are right angles – diagonals are equal in length – diagonals bisect each other – two axes of symmetry – has rotational symmetry Square – opposite sides parallel

– all sides equal

– all angles are right angles – diagonals equal in length

– diagonals bisect each other at right angles – diagonals bisect the angles of the square – four axes of symmetry

– has rotational symmetry Rhombus – opposite sides parallel

– all sides equal – opposite angles equal

– diagonals bisect each other at right angles – diagonals bisect the angles of the rhombus – two axes of symmetry

– has rotational symmetry

Kite – two pairs of equal adjacent sides – one pair of opposite angles equal – one axis of symmetry

– diagonals intersect at right angles

Working mathematically

Reasoning: Is a square a rhombus?

Looking at the definitions of the quadrilaterals above, we see that a parallelogram can also be classified as a trapezium since it has at least one pair of opposite sides parallel. This means that trapeziums are inclusive of parallelograms. Similarly parallelograms are inclusive of rectangles and rhombuses and these are inclusive of squares. This can be represented by a Venn diagram as shown on the next page.

Worksheet 5-02

Naming quadrilaterals

1 a Why is a rectangle a special type of parallelogram but a parallelogram is not always a rectangle? b How can you use the

Venn diagram to answer part a?

2 The kite is not included in the Venn diagram. Where would you put kites on the diagram?

squares rectangles parallelograms trapeziums quadrilaterals rhombuses

Example 2

Prove that the exterior opposite angles of a parallelogram are equal.

Solution

To prove ∠FBC = ∠FDE Proof:

∠FBC = ∠A (corresponding angles, AD || BF)

and ∠FDE = ∠A (corresponding angles, AB || DF)

∴ ∠FBC = ∠FDE

∴ Exterior opposite angles of a parallelogram equal.

A

D F

E

B C

Geometric constructions: Bisectors

The perpendicular bisectors of the sides and angles of a triangle have some interesting properties. You can do these constructions using compasses and a ruler, but it is more fun to use a dynamic geometry program such as Geometer’s Sketchpad or Cabri Geometry.

Construction 1

Step 1: Draw a triangle and label it ABC.

Step 2: Construct the perpendicular bisectors of sides AB and BC. (In Geometer’s Sketchpad you will need to find

the midpoint of the side first.)

Step 3: Construct the point of intersection of the perpendicular

bisectors and label it O.

Step 4: Construct a circle with centre O and a radius OA. Step 5: Construct the perpendicular bisector of side AC.

1 What do you notice about the circle you have drawn? 2 Do all three bisectors meet at point O?

3 Do point O and the circles change in a significant way when you drag a vertex or side of ∆ABC?

C A

B

O

Using technology

Skillsheet 5-03

Geometrical constructions Skillsheet 5-04 Starting Geometer’s Sketchpad Skillsheet 5-05 Starting Cabri Geometry Geometry 5-01 Side bisectors STAGE

5.3

4 The circle is called the circumcircle, and point O is called the circumcentre of the triangle. What does ‘circum’ mean?

5 Can you explain why this construction produces the circumcentre?

Construction 2

Step 1: Draw another triangle and label it PQR. Step 2: Construct the bisectors of ∠PQR and ∠RPQ. Step 3: Construct the point of intersection of the two angle

bisectors and label it O.

Step 4: Use the drawing tool to draw a circle, with centre O,

just touching QR.

Step 5: Construct the bisector ∠QRP.

6 What do you notice about the circle you have drawn? 7 Do all three bisectors meet at point O?

8 Do point O and the circles change in a significant way when you drag a vertex or side of ∆PQR?

9 The circle is called the in-circle, and point O is called the in-centre of the triangle. Explain. 10 Can you explain why this construction produces the in-centre?

Extension

11 Find the radius of the in-circle of the isosceles triangle with sides 10 cm, 10 cm and 4 cm.

O Q P R Geometry 5-02 Angle bisectors

1 Calculate the size of the angle indicated by the pronumeral in each of the following. (Give reasons for your answer.)

a b c d e f y° 80° m° 36° h° x° a° d° 3d°

Exercise 5-01

Example 1

g h i

2 Find the value of the pronumeral in each of the following. (Give reasons for your answers.)

a b c

d e f

3 Find the value of the pronumeral in each of the following, giving reasons for each step:

a b c

d e f

g h i

4 a Find the value of p, q and w in the diagram on the right. (Give reasons.)

b What type of triangle is ∆XYZ? (Give reasons for your answer.) y° 70° p° 130° h° 42° y° 53° k° 127° w° 55° h° 71° f ° 65° a° 134° t° 40° m° y° 58° _d° 74° 120° m° k° w° h° 115° c° 43° a° w° 78° p° 34° A _B C D X h° 32° 78° w° 56° 65° X p° w° q° 70° 110°

5 State whether each of the following is true (T) or false (F): a A rectangle is a parallelogram.

b An equilateral triangle has three axes of symmetry. c The diagonals of a rhombus are equal.

d A scalene triangle has two equal sides. e A square has four axes of symmetry. f Opposite sides of a trapezium are equal.

g The angles opposite the equal sides in an isosceles triangle are equal. h Opposite angles of a parallelogram are equal.

i The diagonals of a square, a rhombus and a rectangle bisect each other at right angles. j A scalene triangle has no equal sides but one axis of symmetry.

k A rectangle and a rhombus both have only two axes of symmetry. l An equilateral triangle is also an isosceles triangle.

m The diagonals of a parallelogram are equal.

n The angles of a square and a rectangle are equal to 90°. 6 KL = ML and MN = MP. Find the value of x, giving

reasons for each step.

7 Segment YP bisects ∠XYW, segment WP bisects ∠TWY and YX || WT.

Prove that ∠YPW = 90°.

8 ∆BDF is an isosceles triangle and BF || CE. Prove that ∆CDE is isosceles.

9 Segment NK bisects ∠HKL. Find the size of ∠NHK, giving reasons. x° K L M N P Y P X W T F E C B D K H N L M 93° 147° Example 2 STAGE

5.3

10 AC || DE, BC = BE and CD = CE. a Find k, giving reasons. b Find m, giving reasons.

11 Find the value of x and y. (Give reasons for your answers.)

a b c

d e f

12 In the figure on the right, AC || ED, AE || BD and BE || CD. Also, CB = CD.

Prove that ∆ABE is an isosceles triangle.

13 In the figure on the right, LP bisects ∠KLN and LM = LN. Prove LP || MN.

14 In the figure on the right, AC = BC and DC = EC. Prove AB || DE.

15 ∆UXY is an equilateral triangle and WX = XU. Prove that ∠WUY is a right angle.

k° m° D E B A C 42° U Z X V Y W y° x° B A C D x° y° H G E F 40° x° y° H K I J 108° y° x° D C B A E 33° 67° x° y° 20° 54° y° x° S R Q P T E A B C D P K L M N B A C D E U X W Y

Properties of convex polygons

The general name for any plane figure bounded by straight sides is a polygon.

A convex polygon has no interior angles greater than 180°. All the diagonals of a convex polygon lie inside the polygon.

In the work that follows, convex polygons will be simply referred to as ‘polygons’.

Regular polygons

A regular polygon has all sides equal and all angles equal. The first eight regular polygons are: 16 In the figure on the right, WY || PQ.

Prove that the angle sum of ∆PQT is 180°.

17 Segment AC bisects ∠FAB and AD bisects ∠HAB. Prove ∠CAD = 90°. W T P Q Y F C B D A H Worksheet 5-04 Angles in polygons Worksheet 5-05

Find the missing angle

Convex polygon Non-convex polygon

3 sides 4 sides 5 sides 6 sides

7 sides 8 sides 9 sides 10 sides

Equilateral triangle Square Pentagon Hexagon

Heptagon Octagon Nonagon Decagon

Canberra: The national capital

Canberra is located in the northern part of the ACT, 300 km south-west of Sydney and 650 km north-east of Melbourne.

The city was designed by an American architect, Walter Burley Griffin, and construction began in 1913. The ‘centre’ of Canberra is based on an equilateral triangle, bounded by the ‘sides’

(Commonwealth Avenue, Kings Avenue and Constitution Avenue). The smaller ‘Parliamentary triangle’ is bounded by

Commonwealth Avenue, Kings Avenue and King Edward Terrace. The axis of symmetry of the triangle runs from Parliament House across Lake Burley Griffin directly along Anzac Parade to the Australian War Memorial.

What other geometrical features can you see in Canberra’s design?

Angle sum of a convex polygon

The size of each angle of a regular polygon with n sides is given by the following formula.

Exterior angle sum of a convex polygon

The angle sum of a convex polygon with n sides is given by: Angle sum = (n − 2) × 180°

Example 3

Find the angle sum of a polygon with 16 sides.

Solution

Angle sum = (16 − 2) × 180° = 14 × 180° = 2520°

One angle = where n is the number of sides

= angle sum n ---, (n–2)×180° n

---Example 4

1 Find the size of an angle of a regular octagon.

Solution

2 The sum of the angles of a regular polygon is 3960°.

a How many sides does the polygon have? b Find the size of each angle.

Solution

Angle sum = (8 − 2) × 180° or = 6 × 180° = 1080° ∴ Each angle = = 135° Each angle = = 135° a (n − 2) × 180 = 3960 n − 2 = n − 2 = 22 n = 24 ∴ Number of sides is 24. b Each angle = = 165° 1080° 8 ---(8–2)×180° 8 ---3960 180 ---3960 24

Example 5

1 The exterior angle of a regular polygon is 18°. How many sides does the polygon have?

Solution

Let n be the number of sides of the polygon. Sum of exterior angles = n × 18° = 360°

18n = 360

n =

= 20 ∴ The polygon has 20 sides.

2 Each interior angle of a regular polygon is 140°. How many sides does the polygon have?

Solution

Method 1:

Exterior angle = 180° − 140° (since exterior angle + interior angle = 180°)

= 40°

Number of exterior angles = 360 ÷ 40 = 9 ∴ The polygon has 9 sides.

Method 2: = 140 = n × 140 (n − 2) × 180 = 140n So: 180n − 360 = 140n 180n = 140n + 360 180n − 140n = 140n + 360 − 140n 40n = 360 ∴ n = 9 360 18 ---(n–2)×180 n ---n (n–2)×180 n ---× 1 1

Exterior angle sum of a convex polygon

Step 1: Use your drawing program to copy the diagram on the right. Step 2: Measure the exterior angles indicated and calculate

their sum.

Step 3: Repeat Steps 1 and 2 for a quadrilateral, a pentagon

and a hexagon. What is the sum of the exterior angles of a convex polygon?

Using technology

Geometry _5-03

Exterior angles of a polygon

1 Use the rule, Angle sum = (n − 2) × 180°, to find the angle sum of a polygon with: a 20 sides b 10 sides c 15 sides d 24 sides e 30 sides f 12 sides 2 Find the number of sides for a polygon that has an angle sum of:

a 4680° b 1800° c 900°

d 6840° e 2160° f 1260°

3 Find the size of each interior angle of a regular:

a pentagon b nonagon c dodecagon (12 sides) 4 Find the size of each interior angle of a regular polygon which has:

a 30 sides b 45 sides c 7 sides d 60 sides

5 Find the number of sides of a regular polygon which has each of its exterior angles equal to:

a 20° b 15° c 45° d 8° e 30°

6 Find the size of each exterior angle of a regular:

a hexagon b 12-sided polygon c 30-sided polygon

7 Find the number of sides of a regular polygon which has each of its interior angles equal to: a 120° b 165° c 172° d 135° e 156° 8 The interior angle of a regular polygon is 11 times the size of an exterior angle. How many

sides has the polygon?

Exercise 5-02

Example 3

Example 4

Example 5

Converting fractions and decimals to percentages

To convert a fraction or decimal to a percentage, multiply it by 100%. 1 Examine these examples:

a = × 100% = = 2 × 20% = 40%

b = × 100% = = 18 × 4% = 72%

c = × 100% = = = 3 × 20% = 60%

d = × 100% = = = 7 × 4% = 28% 2 Now convert these fractions to percentages:

a b c d e f g h i j k l m n o 2 5 --- 2 5 --- 2 5 ---×100% 1 20 18 25 --- 18 25 --- 18 25 ---×100% 1 4 24 40 --- 24 40 --- 24 40 ---×100% 5 3 3 5 ---×100% 1 20 21 75 --- 21 75 --- 21 75 ---×100% 3 4 21 3 ---×4% 1 7 7 10 --- 33 50 --- 27 60 --- 22 25 --- 24 32 ---30 40 --- 60 75 --- 4 5 --- 11 20 --- 28 80 ---15 50 --- 16 20 --- 54 60 --- 18 40 --- 13 25

---Skillbank 5A

SkillTest 5-01 Converting fractions and decimals to percentages

Proving congruent triangles

Congruent figures have exactly the same size and shape. The symbol for ‘is congruent to’ is ≡.

In congruent figures, matching sides are equal and matching angles are equal.

There are four sets of conditions that can be used to determine if two triangles are congruent. These are known as the four tests for congruent triangles.

3 Examine these examples:

a 0.41 = 0.41 × 100% = 0.41 = 41% b 0.08 = 0.08 × 100% = 0.08 = 8% c 0.9 = 0.9 × 100% = 0.90 = 90% d 0.375 = 0.375 × 100% = 0.375 = 37.5% 4 Now convert these decimals to percentages:

a 0.25 b 0.68 c 0.17 d 0.6 e 0.1 f 0.333 g 0.59 h 0.702 i 0.84 j 0.7 k 0.428 l 0.055 m 0.91 n 0.7825 o 0.314 Worksheet 5-06 Congruent triangles proofs

1 Side, Side, Side (SSS)

If three sides of one triangle are respectively equal to three sides of another triangle, then the two triangles are congruent.

2 Side, Angle, Side (SAS)

If two sides and the included angle of one triangle are respectively equal to two sides and the included angle of another triangle, then the two triangles are congruent.

3 Angle, Angle, Side (AAS)

If two angles and one side of one triangle are respectively equal to two angles and the matching side of another triangle, then the two triangles are congruent.

4 Right angle, Hypotenuse, Side (RHS)

If the hypotenuse and a second side of one right-angled triangle are respectively equal to the hypotenuse and a second side of another right-angled triangle, then the two triangles are congruent.

Geometry 5-04

Congruent triangles

Example 6

Which congruence test (SSS, SAS, AAS, or RHS) can be used to prove that each of these pairs of triangles are congruent?

a b

Solution

a AAS b SAS

In the diagram on the right, ∠W = ∠V and TX ⊥ WV. Prove: a ∆WXT ≡ ∆VXT b X is the midpoint of WV.

Solution

a In ∆WXT and ∆VXT ∠W = ∠V (given) ∠WXT = ∠VXT (TX ⊥ WV) TX is common ∴ ∆WXT ≡ ∆VXT (AAS)

b WX = VX (matching sides of congruent triangles)

∴ X is the midpoint of WV 35° 35° 48° 48° 30 mm 30 mm

Example 7

T X W V

1 Test whether each of the following pairs of triangles are congruent. If they are, which test (SSS, SAS, AAS or RHS) would prove them congruent?

a b c d e f 4 cm 4 cm 10 cm 10 cm 8 cm 8 cm 5 cm 5 cm 40° 40° 70° 70° 9 cm 9 cm 7 cm 7 cm 15 cm 15 cm

Exercise 5-03

Example 6 STAGE

5.3

g h i

2 For these pairs of congruent triangles, find the value of the pronumerals:

a b

c d

e f

3 a AB = CB and EB = DB. b QT ⊥ WT, PW ⊥ WT and QW = PT. Prove ∆ABE ≡ ∆CBD. Prove ∆QTW ≡ ∆PWT.

70° 70° 60° 60° 8 cm 8 cm 55° 55° 10 cm 10 cm 40° 40° 58° 58° 31 mm d mm 50° 50° 25° k° w° 25° 12 cm y cm 33° a° p cm 9 cm 35 mm 35 mm 30 mm 30 mm 53 mm 53 mm 109° p° 11 cm 9 cm 9 cm k cm 60° 60°

For Questions 3 and 4 following, identify matching vertices before writing the proof.

A E B D C P W Q T Example 7

c CH || EG, DH || FG and CH = EG. d YX bisects ∠UXW and UX = WX. Prove ∆CDH ≡ ∆EFG. Prove ∆UXY ≡ ∆WXY.

e ABCD is a square and AY = CX. f LM = NP and LP = NM.

Prove ∆ABY ≡ ∆CBX. Prove ∆LMP ≡ ∆NPM.

g O is the centre of the circles. h ∆FGH is an isosceles triangle, so Prove ∆AOB ≡ ∆COD. FH = FG. HN ⊥ FG and GM ⊥ FH.

Prove ∆FHN ≡ ∆FGM.

4 a ∆PQR is isosceles and QA = RB. b TP = XP and AP = CP.

Prove: Prove: i ∆PQA ≡ ∆PRB i ∆TAP ≡ ∆XCP ii ∆PAB is isosceles. ii TA || XC. C D E F G H U X Y W D X C Y A B N L P M B C D A O H M F N G Q A B R P T A X C P STAGE

5.3

Proving properties of triangles and

quadrilaterals

The congruence tests can be used not only to solve numerical problems but also to establish properties of triangles and special quadrilaterals.

c AB || CD and AD || CB.

Prove:

i ∆ABD ≡ ∆CDB ii AD = CB and AB = CD.

d A and B are the centres of two circles which intersect at C and D.

Prove: i ∆ADB ≡ ∆ACB ii AB bisects ∠DAC. D C A B D C B A

e ∠HEF = ∠GFE and EH = FG. Prove:

i ∆HEF ≡ ∆GFE ii ∠EHF = ∠FGE.

f O is the centre of the circle and LT = MN. Prove: i ∆LOT ≡ ∆MON ii ∠LOT ≡ ∠MON. H K E F G T O L N M

g O is the centre of the circle and

OD ⊥ CE. Prove: i ∆OCD ≡ ∆OED ii OD bisects CE. h AB = AD and CB = CD. Prove: i ∆ABC ≡ ∆ADC ii ∠BCA = ∠DCA iii ∆BCY ≡ ∆DCY iv BY = DY. O D C E A C D B Y

Using the word ‘hence’

When the word ‘hence’ is used, it means use or follow on from the previous result. For example, if a question states:

a Prove ∆ABC ≡ ∆DBC b Hence prove ∠A = ∠D,

then part b follows from what was proved in part a and the final statement in the proof would be: ∠A = ∠D (matching angles of congruent triangles)

Example 8

ABCD is a rectangle.

a Prove that ∆ABD ≡ ∆BAC.

b Hence show that the diagonals of a rectangle are equal.

Solution

a In ∆ABD and ∆BAC:

AD = BC (opposite sides of a rectangle)

AB is common.

∠DAB = ∠CBA = 90° (angles in a rectangle)

∴ ∆ABD ≡ ∆BAC (SAS)

b BD = AC (matching sides of congruent triangles)

∴ The diagonals of a rectangle are equal.

D C

A B

Properties of triangles and quadrilaterals

Isosceles triangle 1

Step 1: Use your drawing program to construct isosceles triangle ABC, so that AB = AC.

Step 2: Measure ∠B and ∠C.

1 What can you say about the angles opposite the equal sides?

Step 3: Construct a line from A through T, the midpoint of BC. Step 4: Measure ∠ATB and ∠ATC.

2 What can you say about AT and BC?

Step 5: Measure ∠BAT and ∠CAT.

3 Does AT bisect ∠BAC?

Isosceles triangle 2

Step 1: Use your drawing program to construct isosceles triangle BEW, so that BE = BW.

Step 2: Construct the perpendicular bisector of EW.

4 Which point does the perpendicular bisector pass through? 5 Does the perpendicular bisector of EW bisect ∠EBW?

A B T C B E T W

Using technology

Geometry 5-05 Properties of triangles and quadrilaterals Worksheet 5-03 Investigating the geometrical constructions STAGE

5.3

Rhombus

Step 1: Construct rhombus ABCD with all sides equal. Step 2: Construct the diagonal AC.

Step 3: Construct the perpendicular bisector of AC.

6 Does the perpendicular bisector pass through B and D?

Step 4: Measure ∠ACD, and ∠ACB.

7 Does AC bisect ∠DCB?

8 What properties of a rhombus have you demonstrated with this construction of a rhombus?

9 Construct a parallelogram and, by drawing the diagonals and using the Measure tool, demonstrate the properties of a parallelogram.

D C

B A

1 ∆XYT is isosceles, so TX = TY. TW is drawn so that XW = WY. a Prove ∆TXW ≡ ∆TYW.

b Hence prove that the angles opposite the equal sides of ∆XYT are equal (that is, show ∠X = ∠Y).

2 LMNP is a rhombus (all sides are equal). a Prove ∆LMN ≡ ∆LPN.

b By drawing the diagonal PM, prove ∆PMN ≡ ∆PML.

c Hence prove that the angles of a rhombus are bisected by the diagonals.

3 ABCD is a parallelogram, so AB || DC and

AD || BC.

a Prove ∆ABD ≡ ∆CDB.

b Hence prove that the opposite sides of a parallelogram are equal.

X W Y T L P M N D C

Exercise 5-04

Example 8

4 ∆ABC is an equilateral triangle (AC = AB = BC). CX is drawn so that

AX = BX. Show that each angle is 60°.

a Prove:

i ∆AXC ≡ ∆BXC ii ∠A = ∠B

b Now draw the interval BY, so that CY = AY (as shown). Prove: i ∆AYB ≡ ∆CYB

ii ∠A = ∠C

c Hence calculate the size of an angle in an equilateral triangle.

5 WXYV is a parallelogram, so WX || VY and

WV || XY.

a Prove ∆WXY ≡ ∆YVW.

b By drawing the diagonal XV, prove ∆VWX ≡ ∆XYV. c Hence prove that the opposite angles of a

parallelogram are equal.

6 DEFG is a parallelogram, so DE || FG, DG || EF,

DE = FG, and EF = GD (as proved in Question 3).

a Prove ∆DEX ≡ ∆FGX.

b Hence prove that the diagonals of a parallelogram bisect each other.

7 ∆XYW is an isosceles triangle. The perpendicular

line (the altitude) from the vertex X to side WY, bisects that side. Prove this result by following the steps below:

a Show ∆XWT ≡ ∆XYT. b Hence show WT = YT.

8 ∆ABC is isosceles, so AB = AC. If AX bisects ∠BAC, then AX ⊥ BC. Prove this result for an isosceles triangle by answering the

following:

a Prove ∆AXB ≡ ∆AXC. b Hence show ∠AXB = ∠AXC. c Hence prove AX ⊥ BC

(show ∠AXB = ∠AXC = 90°).

A X B C A B C Y V _Y W X D E G F X X T W Y A X B C STAGE

5.3

9 In ∆TPR, ∠P = ∠R and TX ⊥ PR. a Prove ∆PXT ≡ ∆RXT.

b Hence prove that the sides opposite the equal angles are equal (that is, prove TP = TR).

10 CDEF is a rhombus. a Prove ∆CBF ≡ ∆EBF.

b Explain why the diagonal CE is bisected.

c Explain how the diagonal DF can be shown to be bisected. d i Explain why ∠CBF = ∠EBF.

ii Hence prove FD ⊥ CE.

e What property of a rhombus has been proved? 11 LMNP is a kite, with LM = LP and NM = NP.

a Use congruent triangles to prove: i ∠LMN = ∠LPN

ii ∠MLP and ∠MNP are bisected by the diagonal LN.

b Prove that ∆LMT ≡ ∆LPT and, hence, show that the diagonal MP is bisected at right angles by the diagonal LN.

T X P R F E B D C L N M T P

Working mathematically

Communicating and applying strategies: The converse (true or false)?

The converse is the reverse of a geometrical rule or theorem written in a back-to-front way. The converse of a statement or rule is not necessarily true. Consider the following example: ‘If a quadrilateral is a rectangle, then both pairs of opposite sides are equal.’

The converse is ‘If both pairs of opposite sides of a quadrilateral are equal, then the quadrilateral is a rectangle’. This is false. (The quadrilateral could be a parallelogram.) For the geometrical result, ‘If lines are parallel, then the alternate angles are equal’, the converse is ‘If alternate angles are equal, then the lines are parallel’.

The converse is also true and is often used to test if two lines are parallel (see ‘Start up’, Question 4 on page 183).

1 a For each of the following, write the converse statement and decide whether the converse is also true:

i If two angles are co-interior they have a sum of 180°.

ii If two sides of a triangle are equal, then the angles opposite those sides are equal. iii In a right-angled triangle, the square on the hypotenuse is equal to the sum of the

squares on the other two sides.

iv If the opposite angles of a quadrilateral are equal, then it is a parallelogram. v If a quadrilateral is a square, then the diagonals bisect each other at right angles. vi If the quadrilateral is a rectangle, then the diagonals bisect each other.

b Compare your answers with those of other students.

Tests for quadrilaterals

The properties of special quadrilaterals (such as the parallelogram, rectangle, and so on) can be used as minimum conditions to prove or test whether a given quadrilateral is a parallelogram, rectangle, square or rhombus.

Example 9

ABCD is a quadrilateral on which AD = BC and AB = CD. BD is a

diagonal. Prove that, if the opposite sides of a quadrilateral are equal, then the quadrilateral is a parallelogram.

Solution

In triangles ABD and CDB,

AD = CB (given)

AB = CD (given)

BD is common to both triangles.

∴ ∆ABD ≡ ∆CDB (SSS)

∴ ∠ABD = ∠CDB (matching angles of congruent triangles)

∴ AB || CD (alternate angles ABD and CDB are equal)

Also, ∠ADB = ∠CBD (matching angles of congruent triangles)

∴ AD || CB (alternate angles ADB and CBD are equal)

∴ ABCD is a parallelogram (opposite sides are parallel)

D C B A Worksheet 5-02 Naming quadrilaterals

1 a ABCD is a quadrilateral in which ∠A = ∠C and ∠B = ∠D. Prove that, if the opposite angles of a quadrilateral are equal, then the quadrilateral is a parallelogram.

b LMNP is a quadrilateral in which LM = NP and LM || NP. PM is a diagonal. Prove that, if a pair of opposite sides of a

quadrilateral are equal and parallel, then the quadrilateral is a parallelogram.

c FGHL is a quadrilateral and the diagonals LG and FH bisect

each other. Prove that, if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

D C A B P N L M L H F M G

Exercise 5-05

Example 9 STAGE

5.3

d PQRT is a quadrilateral with all sides equal. PR is a diagonal. Prove that, if the sides of a quadrilateral are equal, then it is a rhombus.

e CDEF is a quadrilateral and the diagonals CE and FD bisect

each other at right angles. Prove that, if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

f WXYV is a quadrilateral. The diagonals WY and VX are

equal and bisect each other. Prove that, if the diagonals of a quadrilateral are equal and bisect each other, then it is a rectangle.

g ABCD is a quadrilateral in which ∠A = ∠B = ∠C = ∠D = 90°. Prove that, if the four angles of a quadrilateral are right angles, then it is a rectangle.

h TWME is a quadrilateral with all sides equal and ∠M = 90°. Prove that, if the sides of a quadrilateral are equal and one angle is a right angle, then it is a square.

i GHKL is a quadrilateral. ∠G = ∠H = ∠K = ∠L = 90° and GH = GL.

Prove that, if the angles of a quadrilateral are right angles and a pair of adjacent sides are equal, then it is a square.

j MNPT is a quadrilateral. The diagonals MP and TN are equal and bisect

each other at right angles. Prove that, if the diagonals of a quadrilateral bisect each other at right angles and are equal, then it is a square.

T R P Q F E C H D V Y W T X D C A B E M T W L K G H T P M X N

2 a ABCD is a parallelogram and BX = DY. Prove: i ∆ABX ≡ ∆CDY

ii AXCY is a parallelogram.

b AECD is a rhombus. AE = EB. Prove: i ∆CBE ≡ ∆DAE

ii BCDE is a parallelogram.

c ABCD is a parallelogram. AP = AS = CQ = CR. Prove:

i RQ = PS and PQ = RS ii PQRS is a parallelogram.

d AC and DB are diameters of concentric circles, centre O. Prove ABCD is a parallelogram.

e PR and SQ are diameters of concentric circles, centre O. TU ⊥ SQ. Prove PQRS is a rhombus.

f DEFG is a rectangle. W, X, Y and Z are the midpoints of

the sides. Prove:

i WXYZ is a parallelogram ii WXYZ is a rhombus. B A Y D X C A E B C D A S Q D R C P B A E B C O D F T U R O P Q S D W Y E X G Z F STAGE

5.3

Summary of tests for quadrilaterals

A quadrilateral is: a a parallelogram if

– both pairs of opposite angles are equal, or – both pairs of opposite sides are equal, or – both pairs of opposite sides are parallel, or

– one pair of opposite sides are equal and parallel, or – the diagonals bisect each other.

b a rectangle if

– all angles are 90°, or

– diagonals are equal and bisect each other. c a rhombus if

– all sides are equal, or

– diagonals bisect each other at right angles. d a square if

– all sides are equal and one angle is 90°, or

– all angles are 90° and two adjacent sides are equal, or – diagonals are equal and bisect each other at right angles.

Working mathematically

Reflecting: Islamic designs

The Islamic rules for art were collected in the Hadith, a ninth-century text. Islamic art derives its unique style from combining the art of the Byzantines, the Copts, the Romans and the Sassanids.

Islam believes in the balance and harmony of all things in existence. One of the vital beliefs is that the totality of things, all good and evil, proceed from the Lord of all being.

There are strict rules against depicting humans or animals, which might result in idol worship, so an art form developed that was based on geometric designs and calligraphy, which are often interwoven.

Geometric patterns appear in architecture and interiors to organise space and to beautify the environment. All patterns

reflect the pure beauty of numbers, considered to be of divine origin. By their very nature, geometric patterns show variation and order as expressions of unity, which is an attribute of God.

Islamic geometric designs are constructed with skillful use of ruler and compasses (skills that you now have). Interactive geometry enables you to be exact with your constructions.

Geometry 5-06

Construction

This Islamic design starts with the drawing of a square.

Step 1: Use axes and a grid to draw a 6 cm

square.

Step 2: Construct a point at the centre of

each side and construct the diagonals of the square.

Step 3: Construct a circle with centre at the

intersection of the two diagonals and with the side midpoints on the circumference.

Step 4: Continue to construct the figure

shown on the right.

1 Are there other patterns you could have drawn using the construction lines? 2 Can you tessellate the pattern you have drawn?

Converting decimals and percentages to fractions

To convert a decimal to fraction form, first count the number of decimal places in the decimal. • If there is one decimal place, then write the decimal part over 10 and simplify.

• If there are two decimal places, then write the decimal part over 100 and simplify. 1 Examine these examples.

a

(two decimal places, two 0s in the denominator)

b

(one decimal place, one 0 in the denominator)

c

d

2 Now convert these decimals to fraction form:

a 0.75 b 0.28 c 0.3 d 0.14 e 0.06 f 0.85 g 0.32 h 0.49 i 0.56 j 0.9 k 0.72 l 0.65 m 0.2 n 0.24 o 0.53 0.35 = 35 100 --- = 7 20 ---7 20 0.8 = 8 10 --- = 4 5 ---4 5 0.64 = 64 100 --- = 16 25 ---16 25 0.22 = 22 100 --- = 11 50 ---11 50

Skillbank 5B

SkillTest 5-02 Converting decimals and percentages to fractions

Properties of

similar figures

Similar figures are the same shape, but not necessarily the same size.

The symbol for ‘is similar to’ is |||. Enlarging or reducing a given figure will always produce a similar figure and the amount by which a figure is enlarged or reduced is called the scale factor or similarity factor.

3 To convert a percentage to a fraction, we write it over 100 and simplify. Examine these examples.

a b

c d

4 Now convert these percentages to fractions:

a 76% b 10% c 80% d 45% e 88% f 56% g 75% h 31% i 68% j 5% k 60% l 54% m 6% n 49% o 82% 26% = 26 100 --- = 13 50 ---13 50 40% = 40 100 --- = 2 5 ---2 5 8% = 8 100 --- = 2 25 ---2 25 95% = 95 100 --- = 19 20 ---19 20 Skillsheet 5-06 Finding sides in similar triangles

Scale factor = image length original length

---If two figures are similar, then: • the matching angles are equal

Example 10

If ∆ABC ||| ∆DEF, find the value of k.

Solution

= (matching sides in the same ratio)

∴ k = × 40 = 50 40 mm 20 mm 25 mm k mm A B C D E F k 40 --- 25 20 ---25 20

---1 Are the figures similar in each of the following pairs?

a b

c d

e f

2 Find the value of the pronumerals in the following pairs of similar figures:

a b c 9 cm 12 cm 6 cm 8 cm 70° 70° 6 cm 4 cm 5 cm 3 cm 10 cm 8 cm _{5 cm} 4 cm 6.4 cm 8 cm 10 cm 4 cm 3 cm 6 cm 5 cm 2 cm 10 cm 4 cm 9 15 27 y k 10 18 12 7 4 6 m

Exercise 5-06

Example 10 Geometry 5-07 Similar triangles

d e f

3 Find the value of the pronumeral in each of the following pairs of similar triangles:

a b

c d

e f

g h

4 Find the similar figures and determine the scale factor: a b x 6 15 9 8 3 15 h 22 8 14 p 12 30 45 d 30° 30° 9 11 6 y 35° 35° k 8 14 18 25 10 35 a m 24 16 20 30 26 26 y 7 ₁₅ 22 e w 70° 70° 22 16 ₁₂ 10 cm 8 cm 60° A 6 cm 7.5 cm 60° B 6.25 cm 5 cm 60° C 18 cm 20 cm 60° D 16 mm 32 mm A 10 mm 20 mm B 13 mm 25 mm C 24 mm D

Tests for similar triangles

There are four conditions or tests for proving that two triangles are similar. 5 ∆ABC is similar to ∆ADE. Find m,

correct to one decimal place.

6 ∆XYW is similar to ∆XZV. Find p, correct to one decimal place.

7 Colin, who is 1.83 m tall, casts a shadow 1.2 m long. At the same time a light tower casts a shadow 14 m long. What is the height of the tower?

8 State whether each of the following is true (T) or false (F). a The angles of similar figures are equal.

b All circles are similar.

c The sides of similar figures are in the same ratio.

d If two triangles have two pairs of matching angles equal, they are similar. e Since the angles of two rectangles are equal, they must be similar. f All squares are similar.

g All equilateral triangles are similar.

h Any two isosceles triangles are always similar.

9 8 6 A B C D E m 5 8 10 V W X _Y Z p 1.2 m 14 m 1.83 m

Tests for similar triangles

1 Side, Side, Side (SSS)

If the three sides of one triangle are proportional to the three sides of another triangle, then the two triangles are similar.

C 5 2 10 4 8 4 F A B _D E Worksheet 5-07 Congruent and similar triangle proofs STAGE

5.3

2 Side, Angle, Side (SAS)

If two sides of one triangle are proportional to two sides of another triangle, and the included angles are equal, then the two triangles are similar.

3 Angle, Angle (AA)

If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

4 Right angle, Hypotenuse, Side (RHS)

If the hypotenuse and a second side of a right-angled triangle are proportional to the hypotenuse and a second side of another right-angled triangle, then the two triangles are similar. C 5 30 50 3 F A B D E C F A B D E 15 6 5 2

Example 11

Prove that the triangles in each of the following pairs are similar:

a b 9 12 6 8 4 6 E C F G T D 50° 9 6 50° 12 R M N W

Solution

a In ∆CDE and ∆FTG: = =

= = (ratios of matching sides)

= = ∴ = =

∴ ∆CDE ||| ∆FTG (matching sides are in the same ratio or SSS)

b In ∆MWN and ∆BPR: = =

= =

∴ =

∠M = ∠B = 50° (included angles are equal)

∴ ∆MWN ||| ∆BPR (two pairs of matching sides are in the same ratio and the included angles are equal or SAS)

Prove ∆ABE ||| ∆ACD.

Solution

In ∆ABE and ∆ACD: ∠A is common

∠AEB = ∠ADC (corresponding angles, BE || CD)

∴ ∆ABE ||| ∠ACD (two pairs of matching angles equal or AA)

a Prove ∆MRP ||| ∆XRW. b Find the value of k.

Solution

a In ∆MRP and ∆XRW:

∠P = ∠W (alternate angles, PM || XW)

∠M = ∠X (alternate angles, PM || XW)

∴ ∆MRP ||| ∆XRW (two pairs of matching angles are equal or AA)

b = (matching sides in similar triangles)

∴ k = 12 × = 4.8 ED GT --- 6 4 --- 3 2 ---CE FG --- 9 6 --- 3 2 ---CD FT --- 12 8 --- 3 2 ---⎭ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎫ ED GT --- CE FG --- CD FT ---MW BP --- 6 8 --- 3 4 ---MN BR --- 9 12 --- 3 4 ---MW BP --- MN BR

---Example 12

A B E D C

Example 13

P M W k X R 12 20 8 k 12 --- 8 20 ---8 20 ---STAGE

5.3

1 Which similarity test (SSS, AA, SAS, RHS) can be used to prove that each of the following pairs of triangles are similar? Where possible state the similarity factor (scale factor) between the triangles.

a b c

d e f

g h i

2 a Prove ∆XYW ||| ∆XLM. b Prove ∆ABC ||| ∆EDC. c Prove ∆PLM ||| ∆PTW.

d Prove ∆TYX ||| ∆KGX. e Prove ∆GMK ||| ∆LHK. f Prove ∆WYZ ||| ∆ZYX.

18 60 40 27 8 6 9 12 10 6 12 8 4.8 9.6 30 16.875 16 9 _42° 75° 63° 42° 30° 30° 2.4 9.6 6 12 3 7.5 18 25 10.8 15 X L Y W M 6 7.5 10 8 E D B C A P L M T W 18 28.8 7.5 12 T X K G M L K H G W X Y Z

Exercise 5-07

Example 11 Example 12

STAGE

5.3

3 a i Prove ∆GKF ||| ∆HKM. b i Prove ∆CDE ||| ∆FGE. ii Find the value of m. ii Find the value of k.

c Prove ∆BCF ||| ∆DCE and, hence, find d Prove ∆AXM ||| ∆AYN and, hence, find the value of x. the length of AX.

e i Prove that ∆TFR ||| ∆TPD. f ABCD is a parallelogram.

ii If FR = 9 cm, DP = 5 cm and i Prove ∆EFC ||| ∆EAB.

TD = 8 cm, find the length of TR. ii Prove ∆EAB ||| ∆AFD. iii Prove ∆EFC ||| ∆AFD. iv If EF = 20 cm, FC = 8 cm and

AF = 16 cm, find the length of AB.

4 In ∆PWT, ∠W = 90° and WN ⊥ PT. a Prove ∆PWN ||| ∆WTN.

b If PN = 4 cm, NT = 9 cm, find the lengths of WN, WP and WT.

5 a Prove that ∆EFG ||| ∆HFE.

b If FG = 16 cm and EF = 20 cm, find the length of FH.

F K H m M G 12 6 8 D C k E G F 16 18 10 F E D B x C 10 6 8 A X Y M N 9 8 15 T P D F R D F E C A B P N W T F _{G H} E Example 13

Proving geometrical results

General geometrical results can be proved by writing a geometrical argument, where reasons are given at each step of the argument. Many of the following proofs involve congruent or similar triangles.

Example 14

AB is the diameter of a circle, with centre O. P is any point on the

circumference. Prove that ∠APB, the angle in a semi-circle, is a right angle.

Solution

Let ∠OAP = x°

∴ ∠OPA = x° (∆OAP is isosceles, equal radii)

∴ ∠BOP = 2x° (exterior angle of ∆OAP)

∴ ∠OPB + ∠OBP = (180 − 2x)° (angle sum of ∆OPB)

But ∠OPB = ∠OBP (∆OPB is isosceles, equal radii)

∴ ∠OPB = (180 − 2x) ÷ 2 = (90 − x)° ∴ ∠APB = ∠OPA + ∠OPB

= x° + (90 − x)° = 90°

∴ The angle in a semi-circle is a right angle.

ABCD is a parallelogram. HG is any interval joining parallel sides AB and DC and passing through the midpoint, T, of diagonal BD.

Prove that the interval through the midpoint of a diagonal of a parallelogram divides opposite sides equally. (That is, prove

DG = BH.)

Solution

Congruent triangles are used to prove this result. In ∆DGT and ∆BHT:

DT = BT (T is the midpoint of BD)

∠DGH = ∠BHT (alternate angles, DC || AB)

∠GDT = ∠HBT (alternate angles, DC || AB)

∴ ∆DGT ≡ ∆BHT (AAS)

∴ DG = BH (matching sides of congruent triangles)

Prove that the interval joining the midpoints of two sides of a triangle is parallel to the third side and is half its length.

A P O B A P O x° B

Example 15

T D A H B G C D E A

Example 16

Worksheet 5-08 Geometrical proofs order activity Worksheet 5-09 Complete the proofs

Pythagoras’ theorem

Pythagoras’ theorem can be stated as follows:

Solution

Similar triangles are used to prove this result. In ∆ADE and ∆ABC:

=

= (since D and E are the midpoints of AB and AC respectively)

∴ = ∠A is common.

∴ ∆ADE ||| ∆ABC (two pairs of matching sides are in the same ratio and the included angles are equal or SAS)

∴ ∠ADE = ∠ABC (matching angles of similar triangles)

∴ DE || BC (corresponding angles proved equal)

Also = (matching sides of similar triangles)

But = ∴ =

∴ DE = × BC

∴ The interval joining the midpoints of two sides of a triangle is parallel to the third side and is half its length.

AD AB --- 1 2 ---AE AC --- 1 2 ---AD AB --- AE AC ---DE BC --- AD AB ---AD AB --- 1 2 ---DE BC --- 1 2 ---1 2

---The square on the hypotenuse of a right-angled triangle is equal to the sum of the squares on the other two sides.

Working mathematically

Applying strategies and reasoning: Proving Pythagoras

There are many proofs of Pythagoras’ theorem and they are based on: • algebra

• geometry

• dissection of areas

The converse of Pythagoras’ theorem

The converse of Pythagoras’ theorem is as follows:

Proof of the converse of Pythagoras’ theorem

Given: ∆ABC where AC2_{= AB}2_{+ BC}2

Aim: To prove ∠ABC = 90°

Construction: Draw ∆PQR where QR = BC,

PQ = AB, and ∠PQR = 90°. Proof: In ∆PQR, PR2_{= PQ}2_{+ QR}2 _{(Pythagoras’ theorem)} = AB2_{+ BC}2 _{(since PQ = AB, QR = BC)} = AC2 _(given) ∴ PR2_{= AC}2 ∴ PR = AC

So, in triangles ABC and PQR,

PQ = AB (by construction)

QR = BC (by construction)

and PR = AC (proved)

∴ ∆PQR ≡ ∆ABC (SSS)

∴ ∠PQR = ∠ABC (corresponding angles of congruent triangles)

∴ ∠ABC = 90° (since ∠PQR = 90°)

1 Proof by dissection

Explain how this dissection proves Pythagoras’ theorem.

2 Find two other proofs of Pythagoras’ theorem by dissection of areas.

A b c a C B a a b b A C B Geometry 5-08 Pythagorean dissection

If the square on one side of a triangle equals the sum of the squares on the other two sides, then the angle between these other two sides is a right angle.

C

A

B R

P

Example 17

1 Determine whether ∆ABC, shown, is right-angled at C.

Solution

202_{+ 99}2_{= 400 + 9801}

= 10 201 = 1012

∴ The angle between AC and BC is a right angle. ∴ ∠C = 90° 2 PQTS is a square with SQ = 10 cm. Find: a the length of PQ b the area of PQTS

Solution

a Let PQ = x cm ∴ PS = x cm

∴ x2_{+ x}2_{= 10}2 _{(by Pythagoras’ theorem)}

∴ 2x2_{= 100} x2_{= 50} x = = ∴ PQ = cm b Area = PQ × PS = = 50 cm2

or since PQTS is a rhombus (why?) Area = = = 50 cm2 20 cm 99 cm 101 cm C B A 10 cm P S T Q 50 5 2 5 2 5 2 ( )×(5 2) 1 2 ---×SQ×PT 1 2 ---×10×10

(Using dynamic geometry software, first confirm each result by construction and measurement before proving the result.)

1 Prove that the exterior angle of a triangle is equal to the sum of the interior opposite angles (that is, prove ∠CBD = ∠CAB + ∠ACB).

A B C D

Exercise 5-08

Example 14 Geometry STAGE

5.3

2 The line joining the centres of two intersecting circles is the perpendicular bisector of the common chord. Prove this result by working through the following:

a Prove that ∆AXB ≡ ∆AYB and, hence, prove that ∠XAC = ∠YAC.

b Prove that ∆XAC ≡ ∆YAC.

c Hence, show that XC = YC and AB ⊥ XY.

3 The line through the midpoint of a side of a triangle parallel to another side bisects the third side. (This result is the converse of the result in Example 16.) Follow the steps below to prove this result.

a In the diagram, Q is the midpoint of PM and QR || MN. Show that ∆PQR ||| ∆PMN.

b Hence, show that PR = RN.

4 In a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. Prove Pythagoras’ theorem using the following steps. a Consider ∆ABC with ∠ABC = 90º and CD ⊥ AB. Show

∆ADC ||| ∆ACB and, hence, that AC2_{= AB × AD (using}

the ratio of matching sides).

b Show ∆BDC ||| ∆BCA and, hence, that BC2_{= AB × DB.}

c Hence, show that AC2_{+ BC}2_{= AB}2_.

5 DEF is a triangle and DG ⊥ EF. Prove, using Pythagoras’ theorem, that:

DE2_{+ GF}2_{= DF}2_{+ EG}2_.

6 Determine whether these are right-angled triangles:

a b c Y C X A B P R N M Q A D C B D E G F 153 135 71 H J I X Y Z 5_ 8 3 _ 8 1 _ 2 2 F D E 2 2 Example 15 Example 16

7 Use Pythagoras’ theorem to calculate the following. (Express your answers in surd form where necessary.)

a Calculate the length of the diagonal of a rectangle with sides 28 cm and 45 cm. b A square has diagonals of length 24 m. Calculate the side length.

c Find the altitudes of an equilateral triangle with sides 10 cm.

d A rhombus has sides of 52 m and one diagonal 40 m. Find the length of the other diagonal. 8 If the diagonals of a quadrilateral are perpendicular then the

sums of the squares on opposite sides are equal. Prove this result by the following steps.

a Use Pythagoras’ theorem to write expressions for AB2_,

BC2_{, CD}2 _{and AD}2_.

b Hence, show AB2_{+ CD}2_{= BC}2_{+ AD}2_.

9 a For the diagram on the right, prove ∆ABC ||| ∆EDC. b Find the value of m.

D A B X C 3 10 5 4 m A B C D E

1 a Use Pythagoras’ theorem to prove that the area of the semi-circle on the hypotenuse of a right-angled triangle is equal to the sum of the areas of the semi-circles on the other two sides.

b Prove that, in any right-angled triangle, the area of the equilateral triangle on the hypotenuse is equal to the sum of the areas of the equilateral triangles on the other two sides.

2 The median is a line joining a vertex of a triangle to the midpoint of the opposite side. Prove that the medians of a triangle are concurrent (that is, they meet at one intersection point). A

C

B

Power plus

Topic overview

• Copy and complete the table below. The best part of this chapter was… The worst part was…

New work…

I need help with…

Language of maths

AA AAS acute-angled angle sum

argument axis of symmetry bisect concurrent congruence test congruent (≡) converse convex deductive diagonal equilateral exterior angle hence hypotenuse interior angle isosceles kite matching midpoint obtuse-angled opposite order parallel parallelogram perpendicular polygon Pythagoras’ theorem quadrilateral ratio reason rectangle regular polygon rhombus RHS right-angled rotational symmetry

SAS scale factor scalene similar (|||) similarity factor similarity test square SSS

test trapezium triangle vertices

1 What is the formal definition of a trapezium?

2 What word in the list above has the same meaning as ‘therefore’? 3 What is the most general name for a shape with four equal sides? 4 What does a test for a quadrilateral do? Give an example of one.

5 Copy and complete: Matching sides in similar figures are in the same _______. 6 Are all squares similar?

7 What does AA mean?

8 Write the converse of Pythagoras’ theorem.

Worksheet 5-10 Geometry crossword Worksheet 5-11 Geometry summary poster ?

• Copy and complete this overview of the topic into your workbook. Add extra words and use pictures to make the overview useful to you. Have your overview checked by other students or by your teacher to make sure nothing is missing or incorrect.

DEDUCTIVE

GEOMETRY

Convex polygons Properties of triangles and quadrilaterals Congruence Similarity • angle sum = 180° or 360° or 540° or … • regular polygons … Scale factor SSS SAS AAS RHS

Tests for congruent triangles

Geometrical proofs using congruence/similarity

Tests for similar triangles: SSS SAS AA RHS c _a b c2 = a2 + b2

1 Find the value of the pronumeral in each of the following. (Give reasons for your answers.)

a b c

2 a In ∆PRT, PR = PT. PR is produced to M so that RM = RT. Prove ∠PTR = 2 × ∠MTR.

b ∆BCD is equilateral, ∆DBA is isosceles with BD = AB. Prove AD ⊥ CD.

3 a Show that the exterior angle of a regular dodecagon (12 sides) is 30°.

b The size of an interior angle in a regular polygon is 175°. How many sides has the polygon?

4 Which congruence test (SSS, SAS, AAS or RHS) can be used to prove that the triangles in each of these pairs are congruent?

a b c

5 Which of the following pairs of triangles are congruent? Give reasons.

a b c

Chapter 5

Review

Ex 5-01 110° c° C B A D F E 114° 4w° w° 36° y° P T R M Ex 5-01 A B D C Ex 5-02 Ex 5-03 5 5 8 8 78° 78° 60° 60° 8 cm _{8 cm} Ex 5-03 Topic test Chapter 5