Chapter 7 Deductive Geometry

Chapter 07 More on Deductive Geometry () New Century Mathematics (Oxford Canotta Maths) Section Topic Teaching Notes Classwork or Homework

7.1 Using the Deductive Approach to Solve Geometric Problems on Triangles

- Students should be able to give formal proof for congruent triangles. - Students should be able to give formal proof for similar triangles. - Students should be able to give formal proof for isosceles triangles.

Ex.7A Q.4-6 Ex.7A Q.7-9 Ex.7A Q.13-15

7.2 Special Lines in Triangles - Students should be able to identify angle bisector ( ), perpendicular bisector (), median () and altitude ().

Ex.7B Q.11-13

7.3 Relations between Lines in a Triangle - Students should be able to use Triangle Inequality (). - Students should be able relate angle bisector with incentre ( ),

perpendicular bisector with circumcentre (), median with centroid () and altitude with orthocentre ().

- Students should be able to recall the properties of the special lines and the special centres of triangles.

Ex.7C Q.2, 7 Ex.7C Q.8-11

Appendix 1. Four centres and two circles of a triangle

Name of centre Construction line Special circle relating the centre Other feature centroid () median () -- centroid divides the line from vertex to midpoint in the ratio 2 : 1 orthocenter () altitude () -- If AH and BK are the lengths of the altitudes, AHBK =

ACBC

incentre () angle bisector () incircle () circumcentre () perpendicular bisector circumcircle ()

2. Positions of the centres in different types of triangles

Centre In acute-angled triangle In right-angled triangle In obtuse-angled triangle

Centroid Inside the triangle Inside the triangle Inside the circle Circumcentre Inside the triangle At the midpoint of the hypotenus Outside the circle Incentre Inside the triangle Inside the circle Inside the circle Orthocentre Inside the triangle At the right-angle vertex Outside the circle

3. For an isosceles triangle with AB = AC, the 4 special lines (altitude, angle bisector, median, perpendicular bisector) through A are the same line. 4. Geometric reasons relating angles and sides:

Condition Conclusion Geometric reason 2 equal sides in a triangle opposite angles are equal s opp. eq. sides an isosceles triangle base angles are equal base s, isos. 2 equal angles in a triangle opposite sides are equal sides opp. eq. s longer side larger greater , greater side larger longer side greater side, greater

5. Triangle Inequality for a triangle with a < b < c, a + b > c

c b < a

a+b = 180o

adjacent angles on straight line

adj.s on st.line

a+b+c = 360o

angles at a point

s at a pt.

a = b

vertically opposite angles

vert.opp. s b = d AB//CD corresponding angles, AB//CD corr. s, AB//CD a = c AB//CD alternate angles, AB//CD alt. s, AB//CD

b+c = 180o

AB//CD interior angles, AB//CD int. s, AB//CD corresponding angles equal corr. s equal alternate angles equal alt. s equal

AB // CD

interior angles supplementary int. s supp.

SSS

SAS

ASA

AAS

RHS

3 sides proportional 3 sides prop.

2 sides in ratio, included angles

equal

2 sides in ratio,

inc. s

equiangular AAA

a+b+c = 180o

angle sum of triangle

sum of

a+b = c

exterior angle of triangle

ext. of

= (n-2)180o

angle sum of polygon

sum of polygon

= 360o

sum of exterior angles of polygon

sum of ext. s of polygon

a = b

base angles of isosceles triangle

base s, isos.

a = b

sides opposite equal angles

sides opp.equal s

Handout 07-1

ABC

AB = AC ABC is an isos. with AB = AC

base angles equal

base s equal

opposite sides equal opp.sides equal opposite angles equal opp.s equal diagonals bisect each other diags.bisect each

other

2 sides equal and parallel 2 sides equal and // opposite sides of parallelogram opposite angles of parallelogram

diagonals of parallelogram

prop.of //gram

property of rectangle prop.of rect. property of rhombus prop.of rhom. property of square prop.of sq.

MN // BC

MN = 12 BC

midpoint theorem

mid-pt.th.

BD = DF

intercept theorem

intercept th.

APPB =

AQQC

equal ratios theorem

equal ratios th.

PQ // BC

converse of equal ratios theorem

converse of eq.ratios

th.

PA = PB

perpendicular bisector theorem

bisector th.

PH = PK

angle bisector theorem

bisector th.

Project 07-1

YAN OI TONG TIN KA PING SECONDARY SCHOOL F.3 Mathematics Challenge Questions Chapter 07 More on Deductive Geometry F.3_____ Name:____________________________________( ) *************************************************************************************** 1. In the figure, an isosceles triangle ABC can be divided into 2 smaller isosceles triangles.

Suggest all other isosceles triangles. Draw these triangles and indicates the sizes of all angles in each triangle.

Example:

2. In a star-shaped figure ABCDE, it is known that B = C = D = E = 34o. Find A.

3. How many isosceles triangles can be found in the following figure?

Chapter 07 More on Deductive Geometry Quiz 07-0 F.3_____ Name:___________________________________________________( ) Marks:_____ / 44

1. Given BAC = CDE. (a) Name a pair of similar triangles

and give reason.

(b) Find the lengths of CE and DE. (6 marks)

2. Given ACB = ECD. (a) Name a pair of similar triangles

and give reason.

(b) Find the lengths of CD and DE. (6 marks)

3. Given ABC = BDC and ACB = CBD. (a) Name a pair of similar triangles

and give reason.

(b) Find the length of BD. (4 marks)

4. Given AXY = ACB. (a) Name a pair of similar triangles

and give reason.

(b) Find the lengths of YC and BC. (6 marks)

(a) ABC ~ DEC (AAA / equiangular)

(b) DE2 =

CE2.5 =

63

CE = 5

DE = 4

E A 5 20

3 B 4 C D (a) ABC ~ EDC

(AAA / equiangular)

(b) CD4 =

DE3 =

205

CD = 16

DE = 12

(a) ABC ~ AYX (AAA / equiangular)

(b) AC2.5 =

BC4 =

2.5 + 3.53

AC = 5

YC = 2

BC = 8

(a) ABC ~ CDB (AAA / equiangular)

(b) BD9 =

96

BD = 13.5

5. Given AB // DE. (a) Name a pair of similar triangles

and give reason.

(b) Find the lengths of CE and DE. (6 marks)

6. Given ACE = 90o. (a) Name a pair of similar triangles

and give reason.

(b) Find the lengths of CD and DE. (6 marks)

7. Given ABC = DBC and ACB = CDB. (a) Name a pair of similar triangles

and give reason.

(b) Find the length of BD. (4 marks)

8. Given XY // BC. (a) Name a pair of similar triangles

and give reason.

(b) Find the lengths of YC and BC. (6 marks)

(a) ABC ~ EDC (AAA / equiangular)

(b) DE2 =

CE3 =

62.5

CE = 7.2

DE = 4.8

E A 5 20

3 B 4 C D (a) ABC ~ CDE

(AAA / equiangular)

(b) CD3 =

DE4 =

205

CD = 12

DE = 16

(a) ABC ~ AXY (AAA / equiangular)

(b) AC3 =

BC4 =

2.5 + 3.52.5

AC = 7.2

YC = 4.2

BC = 9.6

(a) ABC ~ CBD (AAA / equiangular)

(b) BD9 =

94

BD = 20.25

Chapter 07 More on Deductive Geometry Quiz 07-1 F.3_____ Name:_________________________________________________( ) Marks: _____ / 24 1. (a) Write down 2 triangles similar to ABD. (b) Hence, find the length of AC. (4 marks)

2. Given EFG = EGI = GHI= 90o and EGF = EIG

(a) Prove that EGI ~ GHI. (b) If EF = 16 and FG = 12, find the length of HI.

(8 marks)

A

B 8 C 18 D

E I

F G H

(a) ABD ~ CBD ~ CAD (AAA)

(b) Consider CBD and CAD,

AC8 =

18AC (corr. sides, ~s)

AC 2 = 144

AC = 12

A familiar figure:

F.1 Given BC and AC, and ask for CD.

(Similar triangles)

F.2 Given AB and AD, and ask for AC.

(Pythagoras Theorem)

F.3 Given BC and CD, and ask for AC.

(Similar triangles)

Acutally, given any 2 lengths, lengths of the rest can

be found.

(a) In EGI and GHI, 1. EGI = GHI = 90o (given) 2. EGF = EIG (given) 3. EGF + 90o + IGH = 180o

(adj. s on st. line) 4. EIG + 90o + IEG = 180o

( sum of ) 5. ie. IEG = IGH (fr. (2), (3) & (4))

6. EGI ~ GHI (AAA)

(b) In EFG, applying Pythagoras Theorem, EF = 16, FG = 12, EG = 20

As EFG ~ EGI,

EFEG =

FGGI =

EGEI (corr. sides, ~s)

EI = 25, GI = 15

As EGI ~ GHI,

HIGI =

GIEI (corr. sides, ~s)

HI = 9

NB: With the given conditions, the area of trapezium

EFHI can also be found.

3. Find the length of AC. (State the similar triangles you used and no proof is

required in this question.) (5 marks)

4. In the figure, GJ and LI intersect at P. GP = 45, PJ = 36 and GH // LI // KJ.

G

L 45 27

63 P K

36

H I J (a) Prove that GHJ ~ PIJ.

(b) Find the lengths of PI and GL. (7 marks)

In BCF and applying Pythagoras Theorem, BC2 = 602 + 802

BC = 100

As ABC ~ AED (equiangular / AAA), AC

AC + 156 = 100180

AC = 195

NB: BCF ~ DEF and ABC ~ AED, lengths of BC and ED are important.

(a) In GHJ and PIJ, 1. GJH = PJI (common )

2. HGJ = IPJ (corr. s, GH // PI) 3. GHJ = PIJ (corr. s, GH // PI) 4. GHJ ~ PIJ (AAA)

(b) As GHJ ~ PIJ,

PIGH =

JPJG (corr. sides, ~s)

PI = 36 6336 + 45

= 28

Similarly, GPL ~ GJK (AAA)

GLGK =

GPGJ (corr. sides, ~s)

81 GL = 45 (GL + 27)

36 GL = 45 27 GL =

1354

NB: The question can given the lengths of GH, PI and

LP, and calculate the length of KJ.

JPI ~ JGH and GPL ~ GJK, lengths of GP and PJ are important.

Chapter 07 More on Deductive Geometry Quiz 07-2 F.3_____ Name:_________________________________________________( ) Marks: _____ / 21 1. In the figure, AD = CD, AC = BC, AC = 8 cm and

AB // DC. Area of ACD is 12 cm2.

(a) Prove that the two isosceles triangles are

similar. (b) Hence, find the length of AB. (7 marks)

2. In the figure, GEJ is a right-angled triangle, and FHIK is a square, of side length x.

Find the value of x. State the similar triangles you used and no proof is required in this question.

(4 marks)

E F K

G 8 H x I 50 J

(a) Given AD = AC, ACD is an isos.. Given AC = BC, ACB is an isos.. In isos.ACD, 1. DAC = DCA (base s, isos.) 2. DAC + DCA + ADC = 180o ( sum of ) In isos.ACB,

3. CAB = CBA (base s, isos.) 4. CAB + CBA + ACB = 180o

( sum of )

5. DCA = CAB (alt.s, DC // AB) 6. DAC = DCA = CAB = CBA (fr. (1),(3) & (5))

7. ADC = ACB (fr. (1)-(6)) 8. ACD ~ABC (AAA) NB: The underlined 2 points must be corresponding.

(b) Given AC = 8 cm and area of ACD = 12 cm2, height of ACD = 3 cm In ACD, applying Pythagoras Theorem, AD = CD = 5 cm

As ACD ~ABC,

ABAC =

ACAD (corr. sides, ~s)

AB = 12.8 cm

As GFH ~ KJI (AAA)

FHJI =

GHKI (corr. sides, ~s)

x2 = 8 50 x = 20

c.f. Q.1 of Quiz 07-1

3. Given AB = 60, EF = 20 and AB // DC // EF.

A

60 D E 20

B C F

Find the length of CD. State the similar triangles

you used and no proof is required in this question.

(5 marks)

4. In ABC, AB = BC = CD and AD = BD.

Find the size of ACB. State all the geometric reasons you have used. (5 marks)

Let BAC = . In ABD, BAC = ABD = (base s, isos.) In ABC, BAC = ACB = (base s, isos.) BDC = BAD + DAB (ext. of ) = 2 In BCD, BDC = DBC = 2 (base s, isos.) and BCD + BDC + CBD = 180o ( sum of ) 5 = 180o = 36o Hence, ACB = 36o

EITHER Using ABF ~ DCF (AAA)

CFBF =

CD60

Using BCD ~ BFE (AAA)

BCBF =

CD20

NB: Lengths of BC and CF are hidden.

OR Using ABD ~ FED (AAA)

BDED =

6020

Using BCD ~ BFE (AAA)

BDBE =

CD20

NB: Lengths of BD and DE are hidden.

OR Using ABD ~ FED (AAA)

ADFD =

6020

Using CDF ~ BAF (AAA)

DFAF =

CD60

NB: Lengths of AD and DF are hidden.

CD = 15

Chapter 07 More on Deductive Geometry Quiz 07-3 F.3_____ Name:_________________________________________________( ) Marks: _____ / 20 1. In the figure, BPQC is a straight line. BP = CQ and AP = AQ.

(a) Prove that APB = AQC. (b) Prove that APB AQC. (5 marks)

Alternative method for Q.2

Adding the altitude AD in the triangle.

As ABC is an isos. with AB = AC, AD is also a median

area of ABC = 12 (CF)(AB) = 12 (AD)(BC)

ie. CF AB = AD BC CF2 AB2 = AD2 BC2

(BC2 BF2) AB2 = AD2 BC2 BC2 (AB2 AD2) = BF2 AB2

BC2 BD2 = BF2 AB2 14 BC

2 BC2 = BF2 AB2 BC2 = 2 AB BF

2. In the figure, AB = AC and CF is an altitude of ABC.

Prove that BC 2 = 2 AB BF. (9 marks) (Hint: Draw an altitude of ABC through A.)

Construct the altitude AN as shown in the figure. In ABN and ACN, 1. AN = AN (common side) 2. ANB = ANC = 90o (as constructed) 3. AB = AC (given) 4. ABN ACN (SAS) In BCF and BNA, 5. CBF = NBA (common ) 6. BFC = BAN = 90o (given) 7. CBF + BFC + BCF = 180o ( sum of ) 8. NBA + BAN + BNA = 180o ( sum of ) 9. BCF ~ BNA (AAA)

Hence, BCAB =

BF12BC

(corr. sides, ~s)

BC 2 = 2 AB BF

(a) In APQ, 1. AP = AQ (given)

2. APQ = AQP (base s, isos.) 3. APB + APQ = 180o (adj. s on st. line) 4. AQC + AQP = 180o (adj. s on st. line) 5. APB = AQC (fr.(2)-(4))

(b) In APB and AQC, 6. AP = AQ (given)

7. BP = CQ (given)

8. APB = AQC (from (a)) 9. APB AQC (SAS)

3. In the figure, an isosceles triangle ABC can be divided into 2 smaller isosceles triangles.

Suggest all other isosceles triangles.

Draw these triangles and indicates the sizes of all angles in each triangle.

Example: The figure of Q.4 in Quiz 07-2. (6 marks)

[Level 1] Questions of Similar Triangles:

- The triangles are seperated.

- Either 2 angles in each triangles are given,

or a pair of parallel lines are given.

- Example: Q.1-8 in Quiz 07-0, and

Q.4 in Quiz 07-2.


- Pythagoras Theorem is involved.

- Example: Q.1 in Quiz 07-1,

Q.2 in Quiz 07-1, and

Q.1 in Quiz 07-2.


- Pairs of similar triangles cascaded together.

- Usually omitting the lengths of a pair of lines which are

common to 2 paris of similar triangles.

- Example: Q.3 in Quiz 07-1,

Q.4 (modified) in Quiz 07-1, and

Q.3 in Quiz 07-2.

Chapter 07 More on Deductive Geometry Quiz 07-4 F.3_____ Name:_________________________________________________( ) Marks: _____ / 22 In this quiz, write down ALL geometric reasons you have used.

1. In the figure, prove that ABC is an isosceles triangle. (5 marks)

2. The lengths of the 3 sides of an isosceles triangle are 3x cm, (3x + 6) cm and 5x cm. Find (a) the value of x, and (b) the perimeter of the triangle. (4 marks)

3. In the figure, AB = CD, CAB = ECD and ABC = CDE. Which of the following must be true?

c ABC CDE d ABC EAC e EAC is an isosceles triangle A. conly B. eonly C. cand donly D. cand eonly E. c, dand e 4. Which of the following statements about the triangles

in the figure must be true?

A. I and III are similar. B. I and IV are similar. C. II and III are similar. D. II and IV are similar.

As 78o + (3x - 12o) + (2x + 9o) = 180o

( sum of ) x = 21o

ie. The 3 angles of the triangle are 78o, 51o and 51o.

So, ABC is an isosceles triangle with AB = AC. (sides opp. eq. s. / base s eq.)

(a) In an isosceles triangle, there has two equal

sides,

as 3x < 3x + 6 and 3x 5x, then 3x + 6 = 5x

x = 3

(b) Hence, the lengths of the 3 sides are 9 cm,

15 cm and 15 cm respectively.

Perimeter = 39 cm

5. In ABD, C is the mid-point of BD and AB = BC. Prove that 3 AB > AD. (3 marks)

6. Match the special lines of a triangle witn the

corresponding centres. (4 marks)

Angle bisector Incentre Median

Orthocentre

Perpendicular bisector Centroid

Altitude

Circumcentre

7. In the figure, ABC is an acute-angled triangle, AB = AC and D is a point lying on BC such that AD is perpendicular to BC.

Which of the following must be true? c The circumcentre of ABC lies on AD d The orthocentre of ABC lies on AD e The centroid of ABC lies on AD A. c and d only B. c and e only C. d and e only D. c, d and e 8. In ABC is an obtuse-angled triangle, which of the

following points must lie outside ABC? c The centroid of ABC d The circumcentre of ABC e The orthocentre of ABC

A. c and d only B. c and e only C. d and e only D. c, d and e

9. In the figure, ABC and AFED are straight lines. ABF = CDE and BE // CD. Which of the

following triangles are similar? c ABF d AEB e ADC

A. c and d only B. c and e only C. d and e only D. c, d and e 10. In the figure, ABCD is a trapezium. Which of the

following must be true?

B c AED is an equilateral triangle d EBCD is a parallelogram e AB = 2DC

A. conly B. donly C. cand donly D. cand eonly E. c, dand e

By Triangle Inequality,

AB + BD > AD

AB + 2 AB > AD

3 AB > AD

Chapter 07 More on Deductive Geometry Quiz 07-S F.3_____ Name:_________________________________________________( ) Marks: _____ / 1. In the figure, AB // CDE // FG, CD = 8 and FG = 12.

Find (a) the length of AB, and (b) the length of DE.

(State the similar triangles you used and no proof is required in this question.)

(9 marks)

2. In the figure, ADB is a straight line. Prove that AB + 2 CD > AC + BC. (6 marks) In ACD, AD + CD > __________(_________________)

In BCD, BD + CD > __________(_________________)

Hence, ____________________ > ___________

____________________ > ___________

3. The figure shows quadrilateral ABCD, where AB is the longest side and DC is the shortest side. Prove that ADC > ABC. (4 marks)

In ABD,

AB > AD (____________________)

ADB > ABD (____________________)In BCD, BC > CD (____________________)

BDC > CBD (____________________)Hence, ________ + BDC > _______ + CBD

______ > _______

4. How many isosceles triangles can be found in the following figure? Name all of them. (6 marks)

(a) BCD ~ BGF (equiangular / AAA)

BCBG =

812

GCD ~ GBA (equiangular / AAA)

8AB =

CGBG

= 1 BCBG

= 13

AB = 24 (b) FED ~ FAB (equiangular / AAA)

DEAB =

FDFB

= CGBG

DE = 8

AC triangle inequality

BC triangle inequality

AD + CD + BD + CD AC + BC

AB + 2 CD AC + BC

given

greater side, greater

given

greater side, greater ADB ABD ADC ABC

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Chapter 7 Deductive Geometry