HKDSE Exam Series — Master 1, Tackle 5 Hot Question Types for Mathematics (Compulsory Part) (New Syllabus Edition)

Solution Guide

12 Basic Geometry
Conventional Questions
Instant Practice 1
1. (a) ∠STC = ∠BAC (given)
∠SCT = ∠BCA (common)
` !STC ~ !BAC (AA)
2: Any correct proof with correct reasons.
1: Any correct proof without reasons.
SC TC
(b) (i) = (corr. sides, ~ !s)
BC AC
= [1M]

AS + 17 =
AS = 7 cm [1A]
(ii) BC = (43 + 8)2
2

= 2601
AB + AC2 = 452 + (7 + 17)2
2
[1M]
= 2601
= BC2
` !BAC is a right-angled triangle. (converse of Pyth. theorem) [1A]

2. (a) ∠DGE = ∠BGA (vert. opp. ∠s)

∠EDG = ∠ABG (alt. ∠s, DC // AB)
` !DEG ~ !BAG (AA)
2: Any correct proof with correct reasons.
1: Any correct proof without reasons.
(b) ∠EDG = ∠BDC (common)
∠DEG = ∠DBC (given)
∠DGE = 180° – ∠EDG – ∠DEG (∠ sum of !) [1M]
= 180° – ∠BDC – ∠DBC
= ∠DCB (∠ sum of !)
= 90° (prop. of rectangle)
` ∠DGE is a right angle. [1A]
(c) DB = (Pyth. theorem)

=
= 2028 cm
From the result of (b), we have
!DEG ~ !DBC (AAA)
GE DE
= (corr. sides, ~ !s)
CB DB
= [1M]

GE = 125 cm [1A]

Note: Giving reasons is optional.

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 12 Basic Geometry

3. (a) ∠CDE = ∠BEA (given)

∠DCE = ∠EBA = 90 ° (prop. of rectangle)
` !EDC ~ !AEB (AA)
2: Any correct proof with correct reasons.
1: Any correct proof without reasons.
(b) (i) CD = BA (prop. of rectangle)
CD CE
= (corr. sides, ~ !s)
BE BA
CD 64
= [1M]
225 CD
CD2 = 225(64)
CD = 120 cm [1A]
(ii) DE2 = CD2 + CE2 (Pyth. theorem)
2 2
= 120 + 64
= 18 496
AE = EB2 + AB2
2
(Pyth. theorem)
2 2
= 225 + 120
= 65 025
AD2 = (64 + 225)2
= 83 521
DE + AE2 = 18 496 + 65 025
2
[1M]
= 83 521
= AD2
` !ADE is a right-angled triangle with ∠DEA = 90°. [1A]
(converse of Pyth. theorem)
Alternative Method:
∠CDE = ∠BEA (given)
∠DEB = ∠DCE + ∠CDE (ext. ∠ of !) [1M]
∠DEA + ∠BEA = 90° + ∠CDE
` ∠DEA = 90° [1A]

4. (a) ∠ABE = ∠BCF = 90° (prop. of square)

∠AEB = ∠BFC (given)
AB = BC (prop. of square)
` !ABE  !BCF (AAS)
2: Any correct proof with correct reasons.
1: Any correct proof without reasons.
(b) ∠CBF = ∠BAE (corr. ∠s,  !s) [1M]
∠ABE + ∠AEB + ∠BAE = 180° (∠ sum of !)
90° + ∠AEB + ∠CBF = 180°
∠AEB + ∠CBF = 90°
∠BGA = ∠AEB + ∠CBF (ext. ∠ of !) [1M]
= 90°
` !ABG is a right-angled triangle. [1A]

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HKDSE Exam Series — Master 1, Tackle 5 Hot Question Types for Mathematics (Compulsory Part) (New Syllabus Edition) Solution Guide

(c) BE = CF = 65 cm (corr. sides,  !s)

BG = (Pyth. theorem)

= [1M]
= 60 cm [1A]

5. (a) AB = AC (given)
` ∠ABE = ∠ACD (base ∠s, isos. !)
BD = CE (given)
` BD + DE = CE + DE
BE = CD
!BAE  !CAD (SAS)
2: Any correct proof with correct reasons.
1: Any correct proof without reasons.
(b) (i) AD = AE = 136 cm (corr. sides,  !s)
MD = EM = 64 cm (given)
` ∠AMD = 90° (prop. of isos. !)
AM = (Pyth. theorem)

= [1M]
= 120 cm [1A]
(ii) AB = BM2 + AM2
2
(Pyth. theorem)
= (161 + 64)2 + 1202 [1M]
= 65 025
BE = [161 + 2(64)]2
2

= 83 521
AB + AE2 = 65 025 + 1362
2
[1M]
= 83 521
= BE2
` !BAE is a right-angled triangle. (converse of Pyth. theorem) [1A]

6. (a) PM = PN (given)
PR ⊥ MN (given)
∠MPR = ∠NPR (prop. of isos. !)
PR = PR (common)
` !PMR  !PNR (SAS)
2: Any correct proof with correct reasons.
1: Any correct proof without reasons.

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 12 Basic Geometry

(b) (i) MQ = NQ (given)

336
=
2
= 168 cm
QR = (Pyth. theorem)

= [1M]
= 576 cm [1A]
(ii) PQ = (Pyth. theorem)

= [1M]
= 49 cm
PM + MR2 = 1752 + 6002
2
[1M]
= 390 625
PR = (49 + 576)2
2

= 390 625
= PM2 + MR2
` !PMR is a right-angled triangle. (converse of Pyth. theorem) [1A]

Instant Practice 2
1. (a) AB = CD (opp. sides of //gram)
∠BAE = ∠DCE (alt. ∠s, AB // DC)
∠ABE = ∠CDE (alt. ∠s, AB // DC)
` !ABE  !CDE (ASA)
2: Any correct proof with correct reasons.
1: Any correct proof without reasons.
(b) (i) There are 4 pairs of congruent triangles. [1A]
(ii) There are 4 pairs of similar triangles. [1A]

2. (a) AM = BM (given)
∠AMN = ∠ABC = 90° (corr. ∠s, MN // BC)
` ∠AMN = ∠BMN = 90° (adj. ∠s on st. line)
MN = MN (common)
!AMN  !BMN (SAS)
2: Any correct proof with correct reasons.
1: Any correct proof without reasons.
(b) (i) There is 1 pair of congruent triangles. [1A]
(ii) There are 3 pairs of similar triangles. [1A]

3. (a) (i) AB = AC (given)

` ∠DBC = ∠ECB (base ∠s, isos. !)
∠BDC = ∠CEB = 90° (given)
BC = CB (common)
` !BCD  !CBE (AAS)
2: Any correct proof with correct reasons.
1: Any correct proof without reasons.

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HKDSE Exam Series — Master 1, Tackle 5 Hot Question Types for Mathematics (Compulsory Part) (New Syllabus Edition) Solution Guide

(ii) ∠ABE = ∠FBD (common)

∠AEB = ∠FDB = 90 ° (given)
` !ABE ~ !FBD (AA)
2: Any correct proof with correct reasons.
1: Any correct proof without reasons.
(b) There are 3 pairs of congruent triangles. [1A]

4. (a) AD = AC (given)
∠ADE = ∠CAD (alt. ∠s, AC // ED)
∠ACB = ∠CAD (alt. ∠s, AD // BC)
∠ADE = ∠ACB
DE = CB (given)
!ABC  !AED (SAS)
2: Any correct proof with correct reasons.
1: Any correct proof without reasons.
(b) ∠CAB = ∠DAE = 82° (corr. ∠s,  !s)[1M]
∠CAD = ∠ACB = 60° (alt. ∠s, AD // BC)
∠BAE = 360° – 82° – 82° – 60° (∠s at a pt.)
= 136°
Note that AB = AE. (corr. sides,  !s)
∠ABE = ∠AEB (base ∠s isos. !)
∠ABE = 180° – 136° (∠ sum of !)[1M]
2
= 22°[1A]

5. (a) ∠AED = ∠ABC (given)

∠ADE = ∠CAD (alt. ∠s, AC // ED)
∠ACB = ∠CAD (alt. ∠s, AD // BC)
∠ADE = ∠ACB
AD = AC (given)
!ABC  !AED (AAS)
2: Any correct proof with correct reasons.
1: Any correct proof without reasons.
(b) ∠CAB = ∠DAE = 86° (corr. ∠s,  !s)[1M]
∠ACB = 180° – 86° – 34° (∠ sum of !)
= 60°
∠CAD = ∠ACB = 60° (alt. ∠s, AD // BC)
Note that AD = AC.
∠ADC = ∠ACD (base ∠s isos. !)
∠ACD = 180° – 60° (∠ sum of !)[1M]
2
= 60°[1A]

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 12 Basic Geometry

Instant Practice 3
1. ∠SPT = 64° – 36° (ext. ∠ of !) [1M]
= 28°
∠SQR = ∠SPT = 28° (∠s in the same segment) [1A]
∠PQR = 90° (∠ in semi-circle) [1M]
∠PQS = 90° – 28°
= 62°[1A]

2. ∠SQR = 58° – 28° (ext. ∠ of !) [1M]

= 30°
∠SPR = ∠SQR = 30° (∠s in the same segment) [1A]
∠QPS = 90° (∠ in semi-circle) [1M]
∠PSQ =∠PRQ = 28° (∠s in the same segment)
∠PQS = 180° – 90° – 28° (∠ sum of !)
= 62°[1A]

3.
∠QPS = 90° (∠ in semi-circle)
∠PQS = ∠PSQ (base ∠s isos. !)
180° – 90°
∠PSQ = (∠ sum of !)[1M]
2
= 45°[1A]
∠SPT = 72° – 45° (ext. ∠ of !)[1M]
= 27°
∠SQR = ∠SPT = 27° (∠s in the same segment) [1A]

Instant Practice 4
1. ∠CAE = ∠BCA = x (alt. ∠s, BC // AE)
∠ADC = 90° (∠ in semi-circle) [1M]
∠CAE + ∠CDE = 180° (opp. ∠s, cyclic quad.)
` x + 90° + α = 180° [1M]
α = 90° – x [1A]
∠DAE + ∠AED + ∠ADE = 180° (∠ sum of !)
β + 110° + (90° – x) = 180° [1M]
β = x – 20° [1A]

2. Connect CD.
∠ACD = ∠ABD = α (∠s in the same segment) [1A]
∠ADC = 90° (∠ in semi-circle)
∠CAD = 180° – α – 90° (∠ sum of !)
= 90° – α [1A]

∠BAD = ∠BDA (base ∠s, isos. !)
∠BAD + ∠BDA + α = 180° (∠ sum of !) [1M]
α
∠BAD = 90° – [1A]
2
∠BAC = 90 – – (90 – α)
° °

= [1A]

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HKDSE Exam Series — Master 1, Tackle 5 Hot Question Types for Mathematics (Compulsory Part) (New Syllabus Edition) Solution Guide

ce
3. Reflex ∠AOC = 2x (∠ at centre twice ∠ at ) [1M]
∠AOC = 360° – 2x (∠s at a pt.) [1A]
∠ADC = 180° – x (opp. ∠s, cyclic quad.) [1M]
∠BCD + ∠ADC = 180° (int. ∠s, AD // BC)
∠BCD = 180° – (180° – x) [1M]
=x [1A]

4. ∠ABD = ∠ADB = 64° (base ∠s, isos. !)

∠EBD = ∠ABD – ∠ABE
= 64° – 27°
= 37°
∠ECB = ∠ADB = 64° (∠s in the same segment)
∠CEB = ∠CBE (base ∠s, isos. !)
∠CEB + ∠CBE + ∠ECB = 180° (∠ sum of !)
2∠CBE + 64° = 180° [1M]
∠CBE = 58°
∠CBD = ∠CBE – ∠EBD
= 58° – 37° [1M]
= 21°
` ∠DAC = ∠CBD (∠s in the same segment)
= 21° [1A]
∠ABC + ∠ADC = 180° (opp. ∠s, cyclic quad.)
64° + 21° + 64° + ∠BDC = 180° [1M]
∠BDC = 31° [1A]

5. ∠EAB + ∠ABE + ∠AEB = 180° (∠ sum of !)

∠EAB + 68° + 27° = 180°
∠EAB = 85°
 : CD
BC  =3:2
 : BC
` ∠DAC : ∠CAB = CD  (arcs prop. to ∠s at
ce
)
=2:3
∠DAC = ∠EAB 2
2+3
= (85°) 2 [1M]
5
= 34°
∠ACB = ∠DAC + ∠AEB (ext. ∠ of !)
= 34° + 27° [1M]
= 61° [1A]
∠EAB + ∠DCB = 180° (opp. ∠s, cyclic quad.)
85° + ∠DCA + 61° = 180° [1M]
∠DCA = 34° [1A]

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 12 Basic Geometry

Instant Practice 5
1. (a) AD = AB (prop. of square)
∠ADE = ∠ABF = 90° (prop. of square)
DE = BF (given)
` !ADE  !ABF (SAS)
2: Any correct proof with correct reasons.
1: Any correct proof without reasons.
(b) ∠BAF = ∠DAE (corr. ∠s,  !s)
∠EAF = ∠EAB + ∠BAF
= ∠EAB + ∠DAE
= ∠DAB
= 90° (prop. of square) [1M]
AE = AF (corr. sides,  !s)
` !AEF is a right-angled isosceles triangle. [1A]
(c) (i) ∠ECF = ∠GBF = 90° (corr. ∠s, AB // DC)
∠FEC = ∠FGB (corr. ∠s, AB // DC) any 2
∠EFC = ∠GFB (common)
` !ECF ~ !GBF (AA)
DC = BC (prop. of square)
DE = BF = 42 cm (given)
EC CF
= (corr. sides, ~ !s)
GB BF
= [1M]
7(BC – 42) = BC + 42
BC = 56 cm [1A]
(ii) AE = (Pyth. theorem)

=
= 70 cm
AF = AE = 70 cm (proved)
Area of !AEF =

= [1M]
= 2450 cm2 [1A]
Alternative Method:
Area of !AEF
= Area of trapezium ADCF – Area of !ADE – Area of !ECF

= [1M]

= 2450 cm2 [1A]

(iii) Let P be the point on EF such that the distance between P and A is the shortest.
Then PA ⊥ EF.
EF = AF + AE
2 2
(Pyth. theorem)
= 70 + 70
2 2

= cm

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HKDSE Exam Series — Master 1, Tackle 5 Hot Question Types for Mathematics (Compulsory Part) (New Syllabus Edition) Solution Guide

Area of !AEF =

` PA = [1M]

≈ 49.4975 cm

` AH ≥ cm > 49 cm

The claim is disagreed. [1A]

2. (a) ∠ABF = ∠CDE = 90° (prop. of rectangle)

AF = CE (prop. of rhombus)
AB = CD (prop. of rectangle)
` !ABF  !CDE (RHS)
2: Any correct proof with correct reasons.
1: Any correct proof without reasons.
(b) (i) Let H be the point of intersection of EF and AC.

FH = EF = 15 cm (prop. of rhombus)

AH = AC = 20 cm (prop. of rhombus)

∠AHF = 90° (prop. of rhombus)

AF = (Pyth. theorem)

= [1M]
= 25 cm
BF = DE = 7 cm (corr. sides,  !s)
` AB = (Pyth. theorem)

= [1M]
= 24 cm [1A]

(ii) Area of !AFB =

= 84 cm2
Least possible length of BG

= [1M]

= cm (or 6.72 cm) [1A]

3. (a) ∠CAB = ∠ADC (given)

∠BCA = ∠CAD (alt. ∠s, BC // AD)
` !ABC ~ !DCA (AA)
2: Any correct proof with correct reasons.
1: Any correct proof without reasons.

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 12 Basic Geometry

(b) (i) = (corr. sides, ~ !s)

= [1M]

AC = 20 cm [1A]
(ii) AB + BC2 = 122 + 162
2
[1M]
= 400
AC = 20
2 2

= 400
= AB2 + BC2
` !ABC is a right-angled triangle. (converse of Pyth. theorem) [1A]
(iii) Area of !ABC =

= 96 cm2
Shortest distance between B and H

= [1M]

= 9.6 cm < 10 cm
` It is possible that the distance between B and H is less than 10 cm.
The claim is agreed. [1A]

Instant Practice 6
1. (a) E

O
C A
F

B
Connect CE.
∠CEB = ∠CDB = 35° (∠s in the same segment) [1A]
∠AEC = 90° (∠ in semi-circle) [1M]
∠AEB = 90° – 35°
= 55° [1A]
(b) Connect OD.
∠OBD = ∠CDB = 35° (alt. ∠s, CD // BO) [1M]
∠ODB = ∠OBD = 35° (base ∠s, isos. !)
∠DCO = ∠CDO (base ∠s, isos. !)
= 35° + 35° [1M]
= 70°
∠AOD = ∠DCO + ∠CDO (ext. ∠ of !)
= 70° + 70° [1M]
= 140°

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HKDSE Exam Series — Master 1, Tackle 5 Hot Question Types for Mathematics (Compulsory Part) (New Syllabus Edition) Solution Guide

Arc length of AD = 2π(8) [1M]

≈ 19.548 cm
< 20 cm
The arc length of AD is not longer than 20 cm. [1A]

2. (a) ∠CEB = ∠CDB = 30° (∠s in the same segment) [1A]

∠AEC = 30° + 60°
= 90° [1A]
` AC is a diameter. (converse of ∠ in semi-circle) [1A]
(b) Connect OD.
∠ODB = ∠OBD = 40° (base ∠s, isos. !) [1M]
∠DCO = ∠CDO (base ∠s, isos. !)
= 30° + 40° [1M]
= 70°
∠COD = 180° – 70° – 70° (∠ sum of !) [1M]
= 40°

Area of sector COD = π(122) [1M]

≈ 50.265 cm2
> 48 cm2
The area of sector COD is greater than 48 cm2. [1A]

3. (a) ∠DAE = ∠CBE (∠s in the same segment)

∠ADE = ∠BCE (∠s in the same segment) any 2
∠AED = ∠BEC (vert. opp. ∠s)
` !EAD ~ !EBC (AA)
2: Any correct proof with correct reasons.
1: Any correct proof without reasons.
AE DE
(b) (i) = (corr. sides, ~ !s) [1M]
BE CE
=
AE = 21 cm [1A]
(ii) AD2 + AE2 = 282 + 212 [1M]
= 1225
DE2 = 352
= 1225
= AD2 + AE2
` !ADE is a right-angled triangle with ∠DAE = 90°. [1A]
(converse of Pyth. theorem)
(iii) ∠DAE = 90°
` DC is a diameter of the circle. (converse of ∠ in semi-circle)
` DC = (Pyth. theorem)

= [1M]
= 100 cm [1A]

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 12 Basic Geometry

Multiple-choice Questions
Instant Practice 7
1. 180°(n – 2) = 2340° (∠ sum of polygon)
n = 15
A is true.
The number of axes of reflectional symmetry of a regular 15-sided polygon is 15.
B is not true.

Each exterior angle = (sum of ext. ∠s of polygon)

= 24°
C is not true.

Each interior angle =

= 156°
D is not true.
The answer is A.

2. 180°(n – 2) = 144°n (∠ sum of polygon)

36°n = 360°
n = 10
A is not true.
Sum of interior angles = 144°(10)
= 1440°
B is not true.
Number of diagonals of a regular 10-sided polygon

= 35
C is not true.
The number axes of reflectional symmetry of a regular 10-sided polygon is 10.
D is true.
The answer is D.

3. Let x be the size of an exterior angle of the polygon.

Then an interior angle of the polygon is 180° – x.
(adj. ∠s on st. line)
180 – x – x = 135
° °
45° = 2x
x = 22.5°
` Each exterior angle of the polygon = 22.5°

n= (sum of ext. ∠s of polygon)

= 16
` I is true.
The number of axes of reflectional symmetry of a regular 16-sided polygon is 16.
` II is not true.
The interior angle of the polygon = 180° – 22.5° = 157.5°.
` III is true.
The answer is B.

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HKDSE Exam Series — Master 1, Tackle 5 Hot Question Types for Mathematics (Compulsory Part) (New Syllabus Edition) Solution Guide

4. Let x be the size of an exterior angle of the polygon.

Then an interior angle of the polygon is 180° – x.
(adj. ∠s on st. line)
180° – x = 9x
x = 18°
Each exterior angle of the polygon = 18°
Each interior angle of the polygon = 180° – 18°
= 162°
` I is not true.

n= (sum of ext. ∠s of polygon)

= 20
The number of diagonals of a regular 20-sided polygon

= 170
` II is not true.
The sum of the interior angles of the polygon = (20 – 2)180° = 3240°.
` III is true.
The answer is B.

5. Let x be the size of an exterior angle of the polygon.

Then an interior angle of the polygon is 180° – x.
(adj. ∠s on st. line)
180° – x = x + 108°
x = 36°
Each exterior angle of the polygon = 36°
Each interior angle of the polygon = 180° – 36°
= 144°
` II is true.

n= (sum of ext. ∠s of polygon)

= 10
` I is true.
The number of axes of reflectional symmetry of a regular 10-sided polygon is 10.
` III is true.
The answer is D.

Instant Practice 8
1. BE is a perpendicular bisector of AD. (given)
` AE = DE and ∠AEB = ∠DEB = 90°
BE = BE (common)
` !ABE  !DBE (SAS)
II is true.
∠BAE = 2∠ABE (given)
∠BAE + ∠ABE + ∠ AEB = 180° (∠ sum of !)
3∠ABE + 90° = 180°
∠ABE = 30°

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 12 Basic Geometry

∠ABE = ∠DBE (corr. ∠s,  !s)

` ∠ABD = 2∠ABE = 60°
I is not true.
∠BAE = 2∠ABE = 60°
∠BDE = ∠BAE = 60° (corr. ∠s,  !s)
` !ABD is an equilateral triangle and AB = BD = AD.
(prop. of equilateral !)
AD = BC (opp. sides of //gram)
` AB = BD = BC
III is true.
The answer is C.

2. AE = BE (given)
∠EAD = ∠EBC = 90° (prop. of rectangle)
AD = BC (prop. of rectangle)
` !AED  !BEC (SAS)
I is true.
∠AED = ∠BEC (corr. ∠s,  !s)
∠AED + ∠CED + ∠BEC = 180° (adj. ∠s on st. line)
2∠AED + 90° = 180°
∠AED = 45°
∠ADE = 180° – ∠EAD – ∠AED (∠ sum of !)
= 180° – 90° – 45°
= 45°
III is not true.
∠ADE = ∠AED = 45°
` AE = AD (sides opp. equal ∠s)
II is true.
The answer is A.

3. AE = BE (given)
∠AED = ∠BED = 90° (given)
DE = DE (common)
` !AED  !BED (SAS)
I is true.
With similar arguments, we have
!BDF  !CDF (SAS)
AD = BD (corr. sides,  !s)
BD = CD (corr. sides,  !s)
` AD = CD
II is true.
∠EAD = ∠EBD (corr. ∠s,  !s)
∠EAD = ∠DCF (opp. ∠s of // gram)
∠DCF = ∠DBF (corr. ∠s,  !s)
` ∠EAD = ∠EBD = ∠DBF
∠EAD + ∠ABC = 180° (int. ∠s, AD // BC)

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HKDSE Exam Series — Master 1, Tackle 5 Hot Question Types for Mathematics (Compulsory Part) (New Syllabus Edition) Solution Guide

∠EAD + ∠EBD + ∠DBF = 180°

3∠EAD = 180°
∠EAD = 60°
` ∠ABC = 180° – ∠EAD
= 120°
III is not true.
The answer is A.

4. Let ∠DBC = x.
Then ∠ABD = x and ∠BAC = x.
∠ACB = ∠ABC = 2x (base ∠s, isos. !)
∠BAC + ∠ABC + ∠ACB = 180° (∠ sum of !)
x + 2x + 2x = 180°
x = 36°
∠BAC = x = 36°
` I is not true.
∠BDC = 180° – ∠DBC – ∠ACB (∠ sum of !)
= 180° – 36° – 2(36°)
= 72°
∠BAC = ∠CBD = 36°
∠ABC = ∠BCD = 72°
∠ACB = ∠BDC = 72°
!ABC ~ !BCD (AAA)
` II is true.
∠BDC = ∠BCD = 72°
` BC = BD (sides opp. equal ∠s)
∠BAC = ∠ABD = 36°
` AD = BD (sides opp. equal ∠s)
` AD = BC
III is true.
The answer is D.

5. ∠EAB = ∠ABC

= (∠ sum of polygon)

= 108°
AE = AB = BC
` ∠AEB = ∠ABE (base ∠s, isos. !)
∠EAB + ∠AEB + ∠ABE = 180° (∠ sum of !)
108° + 2∠AEB = 180°
∠AEB = 36°
Similarly, we have
∠AEB = ∠ABE = ∠BAC = ∠BCA = 36°
∠EAF = ∠EAB – ∠BAC
= 108° – 36°
= 72°

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 12 Basic Geometry

∠AFE = ∠BAC + ∠ABE (ext. ∠ of !)

= 2(36°)
= 72°
= ∠EAF
` AE = EF (sides opp. equal ∠s)
I is true.
With similar arguments, we have
∠ECD = 36°
∠ACE = ∠BCD – ∠BCA – ∠ECD
= 108° – 36° – 36°
= 36°
= ∠BAC
` AB // EC (alt. ∠s equal)
II is true.
∠ACE = ∠AEF = 36°
∠CAE = ∠EAF = 72°
∠AEC = 180° – ∠ACE – ∠CAE (∠ sum of !)
= 180° – 36° – 72°
= 72°
= ∠AFE
` !ACE ~ !AEF (AAA)
III is true.
The answer is D.

6. ∠FAB = ∠ABC

= (∠ sum of polygon)

= 120°
FA = AB = BC
` ∠AFB = ∠ABF (base ∠s, isos. !)
∠FAB + ∠AFB + ∠ABF = 180° (∠ sum of !)
120° + 2∠ABF = 180°
∠ABF = 30°
Similarly, we have
∠BAC = ∠ABF = 30°
` AG = BG (sides opp. equal ∠s)
I is true.
∠FAG = ∠FAB – ∠BAC = 90°
∠CBG = ∠ABC – ∠ABF = 90°
AG = BG (proved)
∠AGF = ∠BGC (vert. opp. ∠s)
` !AGF  !BGC (ASA)
II is true.
EF = AF

= cos ∠AFB

` = cos 30° =

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GF = ≠ EF

III is not true.

The answer is A.

7. ∠AEF = ∠CFE = 90° (given) A D

AE // CF (alt. ∠s equal)
F
I is true.
Join AF and CE.
E
∠ABE = ∠CDF (alt. ∠s, AB // DC)
DF = CD cos ∠CDF = AB cos ∠ABE = BE B C
AD = CB (prop. of rectangle)
∠ADF = ∠CBE (alt. ∠s, AD // BC)
!AFD  !CEB (SAS)
II is true.
AF = CE (corr. sides  !s)
AB = CD (prop. of rectangle)
BF = BE + EF = DF + EF = DE
!ABF  !CDE (SSS)
III is true.
The answer is D.

8. EF = CE (prop. of equilateral !)
CE = AE = DE (prop. of rectangle)
E is the centre of the circle passing through A, D, F and C.
AC is a diameter of the circle.
∠AFC = 90° (∠ in semi-circle)
∴ AF ⊥ FC
I is true.
∠FAC = ∠FDC (∠s in the same segment)
II is true.
∠ADG = ∠CFG = 90°
∠AGD = ∠CGF (vert. opp. ∠s)
∠DAG = ∠FCG (∠s in the same segment)
!ADG ~ !CFG (AAA)
III is true.
The answer is D.

Instant Practice 9
1. ∠ADB = ∠ADC (prop. of rhombus)

= 24°
DE = DB (given)
` ∠DEB = ∠DBE (base ∠s, isos. !)
∠DEB + ∠DBE + ∠ADB = 180° (∠ sum of !)
2∠DEB + 24° = 180°
∠DEB = 78°
i.e. ∠AEB = 78°
The answer is D.
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 12 Basic Geometry

2. EF = EC (given)
` ∠EFC = ∠ECF (base ∠s, isos. !)
∠EFC + ∠ECF + ∠FEC = 180° (∠ sum of !)
2∠ECF + 76° = 180°
∠ECF = 52°
∠ADC = ∠ECF = 52° (alt. ∠s, AD // BC)
AD = FD (given)
` ∠DFA = ∠FAD (base ∠s, isos. !)
∠DFA + ∠FAD = ∠ADC (ext. ∠ of !)
` 2∠FAD = 52°
∠FAD = 26°
The answer is A.

3. ∠EDB = ∠DBC + ∠DFC (ext. ∠ of !)

= 62° + 26°
= 88°
ED = BD (given)
` ∠DEB = ∠DBE (base ∠s, isos. !)
∠DEB + ∠DBE + ∠EDB = 180° (∠ sum of !)
2∠DBE + 88° = 180°
∠DBE = 46°
∠DCF = ∠ABC (corr. ∠s, AB // DC)
= ∠DBE + ∠DBC
= 46° + 62°
= 108°
The answer is B.

4. ∠ADE = ∠EBF = ∠ADC (prop. of rhombus)

= 42°
DE = AD (given)
` ∠DAE = ∠DEA (base ∠s, isos. !)
∠DAE + ∠DEA + ∠ADE = 180° (∠ sum of !)
2∠DEA + 42° = 180°
∠DEA = 69°
EF = BE (given)
` ∠EFB = ∠EBF = 42° (base ∠s, isos. !)
∠DEF = ∠EFB + ∠EBF (ext. ∠ of !)
= 42° + 42°
= 84°
` ∠AEF = ∠DEA + ∠DEF
= 69° + 84°
= 153°
The answer is B.

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HKDSE Exam Series — Master 1, Tackle 5 Hot Question Types for Mathematics (Compulsory Part) (New Syllabus Edition) Solution Guide

5.∠BAD = ∠ACE = 60° (prop. of equilateral !)

BA = AC (prop. of equilateral !)
AD = CE (given)
!CAE  !ABD (SAS)
∠AEC = ∠BDA (corr. ∠s,  !s)
∠BDA = ∠CBD + ∠ACE (ext. ∠ of !)
= 32° + 60°
= 92°
∴ ∠AEC = 92°
∠AEB = 180° – 92° (adj. ∠s on st. line)
= 88°
The answer is C.

6. ∠AGD = 180° – 25° – 90° – 32° = 33° (∠ sum of !)

AD = AB (prop. of square)
AG = AE (prop. of square)
∠DAG = 90° + 25° = 115°
∠BAE = 90° + 25° = 115°
∴ ∠DAG = ∠BAE
!ADG  !ABE (SAS)
∠AEB = ∠AGD = 33° (corr. ∠s,  !s)
∠BEF = 90° – 33°
= 57°
The answer is A.

Instant Practice 10
1. CN = (Pyth. theorem)

=
= 22 cm
AN = (Pyth. theorem)

=
= 15 cm
BC = (Pyth. theorem)

=
= 35 cm
` Area of !ABC =

= 210 cm2
The answer is C.

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 12 Basic Geometry

2. AC = (Pyth. theorem)

=
= 25 cm
AB = (Pyth. theorem)

=
= 312 cm
Area of quadrilateral ABCD
= Area of !ACD + Area of !ABC

= 3984 cm2
The answer is C.

3. DA2 + DB2 = 182 + 242

= 900
AB = 30
2 2

= 900
= DA2 + DB2
` ∠ADB = 90° (converse of Pyth. theorem)
CD = (Pyth. theorem)

=
= 45 cm
The answer is D.

4. Horizontal distance between A and H

= 1 – 5 + 11
=7
Vertical distance between A and H
= –2 + 3 – 5 + 28
= 24
` AH = (Pyth. theorem)
= 25
The answer is A.

5. Horizontal distance between A and G

= –2 + 7 + 3
=8
Vertical distance between A and G
=1+3+2
=6
` AG = (Pyth. theorem)
= 10
The answer is D.

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6. GC = (Pyth. theorem)
= 39 cm
CH = (Pyth. theorem)
= 17 cm
` Perimeter of !HCG
= 39 + 17 + (18 + 18 + 8)
= 100 cm
The answer is B.

Instant Practice 11
1. FB = (Pyth. theorem)

=
= 56 cm
∠EFB = ∠EDA (corr. ∠s, BF // AD)
∠FBE = ∠DAE = 90° (prop. of square)
∠FEB = ∠DEA (common)
` !EFB ~ !EDA (AAA)
Let x cm be the side length of the square ABCD.

= (corr. sides, ~ !s)

105x = 56(105 + x)
x = 120
Side length of the square = 120 cm
DF = (Pyth. theorem)

=
= 136 cm
The answer is D.

2. FB = (Pyth. theorem)

=
= 119 cm
∠EFB = ∠EDA (corr. ∠s, BF // AD)
∠FBE = ∠DAE = 90° (prop. of square)
∠FEB = ∠DEA (common)
` !EFB ~ !EDA (AAA)
Let x cm be the side length of the square ABCD.

= (corr. sides, ~ !s)

408x = 119(408 + x)
x = 168
102 Side length of the square = 168 cm
 12 Basic Geometry

DF = (Pyth. theorem)

=
= 175 cm
The answer is D.

3. ∠DCE = ∠EBA = 90° (prop. of rectangle)

∠DEB = ∠DCE + ∠EDC (ext. ∠ of !)
` 90° + ∠AEB = 90° + ∠EDC
∠EDC = ∠AEB
∠DEC = 180° – ∠DCE – ∠EDC (∠ sum of !)
= 180° – ∠EBA – ∠AEB
= ∠EAB (∠ sum of !)
` !CDE ~ !BEA (AAA)
CD = BA (prop. of rectangle)

= (corr. sides, ~ !s)

CD = 36 cm
DE = (Pyth. theorem)

=
= 45 cm
` Perimeter of !CDE = 36 + 27 + 45
= 108 cm
The answer is C.

4. ∠DCE = ∠EBA = 90° (prop. of rectangle)

∠DEB = ∠DCE + ∠EDC (ext. ∠ of !)
` 90° + ∠AEB = 90° + ∠EDC
∠EDC = ∠AEB
∠DEC = 180° – ∠DCE – ∠EDC (∠ sum of !)
= 180° – ∠EBA – ∠AEB
= ∠EAB (∠ sum of !)
` !CDE ~ !BEA (AAA)
CD = BA (prop. of rectangle)

= (corr. sides, ~ !s)

CD = 60 cm
DE = (Pyth. theorem)

=
= 156 cm
` Perimeter of !CDE = 60 + 144 + 156
= 360 cm
The answer is B.
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HKDSE Exam Series — Master 1, Tackle 5 Hot Question Types for Mathematics (Compulsory Part) (New Syllabus Edition) Solution Guide

5. ∠ABD = ∠DCE = 60° (prop. of equilateral !)

∠DAB + ∠ABD = ∠ADC (ext. ∠ of !)
` ∠DAB + 60° = 60° + ∠EDC
∠DAB = ∠EDC
∠ADB = 180° – ∠ABD – ∠DAB (∠ sum of !)
= 180° – ∠DCE – ∠EDC
= ∠DEC (∠ sum of !)
` !ADB ~ !DEC (AAA)
AB = BC = 36 cm (prop. of equilateral !)
= (corr. sides, ~ !s)

EC = 5 cm
` AE = 36 – 5
= 31 cm
The answer is B.

6. ∠ABD = ∠BAC = ∠AEF = ∠EAD = 60° (prop. of equilateral !)

` ∠BAD + ∠DAF = ∠DAF + ∠EAF
∠BAD = ∠EAF
∠ADB = 180° – ∠ABD – ∠BAD (∠ sum of !)
= 180° – ∠AEF – ∠EAF
= ∠AFE (∠ sum of !)
` !ABD ~ !AEF (AAA)
= (corr. sides, ~ !s)

AF = 49 cm
The answer is C.

7. BE = (Pyth. theorem)

=
= 27 cm
∠ABE = ∠ECF = 90° (prop. of square)
∠BAE + ∠ABE = ∠AEC (ext. ∠ of !)
∠BAE + 90° = 90° + ∠CEF
` ∠BAE = ∠CEF
∠AEB = 180° – ∠ABE – ∠BAE (∠ sum of !)
= 180° – ∠ECF – ∠CEF
= ∠EFC (∠ sum of !)
` !ABE ~ !ECF (AAA)

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 12 Basic Geometry

= (corr. sides, ~ !s)

EF = 11.25 cm
The answer is D.

8. ∠ABE = ∠ECF = 90° (prop. of square)

∠BAE + ∠ABE = ∠AEC (ext. ∠ of !)
∠BAE + 90° = 90° + ∠CEF
` ∠BAE = ∠CEF
∠AEB = 180° – ∠ABE – ∠BAE (∠ sum of !)
= 180° – ∠ECF – ∠CEF
= ∠EFC (∠ sum of !)
` !ABE ~ !ECF (AAA)
= (corr. sides, ~ !s)

CF = 35 cm
` EF = (Pyth. theorem)

= 91 cm
The answer is D.

Instant Practice 12
1. AB = AD (prop. of square)
∠ABF = ∠ADE = 90° (prop. of square)
BF = DE (given)
` !ABF  !ADE (SAS)
AF = AE (corr. sides,  !s)
∠DAE = 180° – ∠ADE – ∠AED (∠ sum of !)
= 180° – 90° – 66°
= 24°
∠BAF = ∠DAE = 24° (corr. ∠s,  !s)
∠EAG = 180° – ∠AED – ∠AGD (∠ sum of !)
= 180° – 66° – 69°
= 45°
∠FAE = ∠BAD + ∠DAE – ∠BAF
= ∠BAD
= 90°
` ∠FAG = ∠FAE – ∠EAG
= 90° – 45°
= 45°
= ∠EAG
AG = AG (common)
` !FAG  !EAG (SAS)

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HKDSE Exam Series — Master 1, Tackle 5 Hot Question Types for Mathematics (Compulsory Part) (New Syllabus Edition) Solution Guide

∠AGF = ∠AGE = 69° (corr. ∠s,  !s)

` ∠CGF = 180° – ∠AGF – ∠AGE (adj. ∠s on st. line)
= 180° – 69° – 69°
= 42°
The answer is A.

2. ∠BCD = ∠BAD = 112° (prop. of rhombus)

CD = CB (prop. of rhombus)
∠CDE = ∠CBF (prop. of rhombus)
DE = BF (given)
` !CDE  !CBF (SAS)
∠ECD = ∠FCB (corr. ∠s,  !s)
∠ECD + ∠FCE + ∠FCB = 112°
` 2∠ECD + 40° = 112°
∠ECD = 36°
The answer is B.

3. BC = AC (prop. of equilateral !)
∠ACB = ∠DCE = ∠DEC = 60° (prop. of equilateral !)
∠ACB + ∠ACD = ∠ACD + ∠DCE
` ∠BCD = ∠ACE
CD = CE (prop. of equilateral !)
` !BCD  !ACE (SAS)
∠BDC = ∠AEC = 60° (corr. ∠s,  !s)
` ∠ADB = 180° – ∠BDC – ∠CDE (adj. ∠s on st. line)
= 180° – 60° – 60°
= 60°
The answer is B.

4. ∠ABC = ∠ACB = ∠BAC = 60° (prop. of equilateral !)

∠ADF = ∠ABC = 60° (corr. ∠s, DE // BC)
∠AED = ∠ACB = 60° (corr. ∠s, DE // BC)
` !ADE is an equilateral triangle and DA = AE = DE.
(prop. of equilateral !)
∠DFA + ∠FAE = ∠AED (ext. ∠ of !)
∠DFA + 23° = 60°
∠DFA = 37°
∠BAE = ∠FDA = 60°
EF = EC (given)
` EF + DE = EC + AE
DF = AC = AB
` !BAE  !FDA (SAS)
∠ABE = ∠DFA = 37° (corr. ∠s,  !s)
∠EBC = ∠ABC – ∠ABE
= 60° – 37°
= 23°
The answer is A.

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 12 Basic Geometry

Instant Practice 13
1. AE = 3692 + 4922 (Pyth. theorem)
= 615 cm
Since ABED is a cyclic quadrilateral,
∠ADE = ∠ABC = 90°. (opp ∠s, cyclic quad.)
ED = 615 – 561
2 2
(Pyth. theorem)
= 252 cm
!ABC ~ !EDC (AAA)
AB AC
= (corr. sides, ~ !s)
ED EC
369 561 + CD
=
252 EC
41EC = 28(561 + CD)
41EC – 28CD = 15 708…… (1)
AB BC
= (corr. sides, ~ !s)
ED CD
369 492 + EC
=
252 CD
41CD = 28(492 + EC)
41CD – 28EC = 13 776 …… (2)
Solving (1) and (2),
CD = 1120 cm
The answer is C.

2. AE = 5802 + 4352 (Pyth. theorem)

= 725 cm
Since ABED is a cyclic quadrilateral,
∠ADE = ∠ABC = 90°. (opp ∠s, cyclic quad.)
ED = 725 – 715
2 2
(Pyth. theorem)
= 120 cm
!ABC ~ !EDC (AAA)
AB AC
= (corr. sides, ~ !s)
ED EC
580 715 + CD
=
120 EC
29EC = 6(715 + CD)
29EC – 6CD = 4290…… (1)
AB BC
= (corr. sides, ~ !s)
ED CD
580 435 + EC
=
120 CD
29CD = 6(435 + EC)
29CD – 6EC = 2610 …… (2)
Solving (1) and (2),
CD = 126 cm
The answer is B.

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HKDSE Exam Series — Master 1, Tackle 5 Hot Question Types for Mathematics (Compulsory Part) (New Syllabus Edition) Solution Guide

3. AE = 4442 + 5922 (Pyth. theorem)

= 740 cm
Since ABED is a cyclic quadrilateral,
∠ADE = ∠ABC = 90°. (opp ∠s, cyclic quad.)
ED = 740 – 704
2 2
(Pyth. theorem)
= 228 cm
!ABC ~ !EDC (AAA)
AB AC
= (corr. sides, ~ !s)
ED EC
444 704 + CD
=
228 EC
37EC = 19(704 + CD)
37EC – 19CD = 13 376…… (1)
AB BC
= (corr. sides, ~ !s)
ED CD
444 592 + EC
=
228 CD
37CD = 19(592 + EC)
37CD – 19EC = 11 248 …… (2)
Solving (1) and (2),
CD = 665 cm
The answer is D.

Instant Practice 14
1. ∠CBE = 60° (prop. of equilateral !)
∠ABE = 90° – 60° = 30°
∠CAB = 45° (prop. of square)
∠CFB = 45° + 30° = 75° (ext. ∠ of !)
I is true.
AB = EB (given)
∠BAE = ∠BEA (base ∠s, isos. !)
180° – 30°
∠BEA = (∠ sum !)
2
= 75°
∠AFE = 75° (vert. opp. ∠s)
∴ AE = AF (sides opp. equal ∠s)
II is true.
!CDE ~ !AFE (AAA)
EF FA
= (corr. sides, ~ !s)
ED DC
III is true.
The answer is D.

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2. !CAE  !CBD (SAS)

∠CAE = ∠CBD (corr. ∠s,  !s)
I is true.
AE = BD (corr. sides,  !s)
II is true.
∠AFB = 180° – ∠ABF – ∠BAF (∠ sum of !)
= 180° – ∠ABC – ∠CBD – ∠BAF
= 180° – ∠ABC – ∠CAE – ∠BAF
= 180° – ∠ABC – ∠BAC
= ∠ACB (∠ sum of !)
= 60°
∠BFE = 180° – 60° (adj. ∠s on st. line)
= 120°
III is true.
The answer is D.

3. !ADG  !ABE (SAS)

∠AGD = ∠AEB (corr. ∠s,  !s)
I is true.
DG = BE (corr. sides,  !s)
II is true.
Let AB and DG meet at I.
Consider !HBI and !ADI.
∠HBI = ∠ADI (corr. ∠s,  !s)
∠BIH = ∠AID (vert. opp. ∠s)
∠BHI = 180° – ∠HBI – ∠BIH (∠ sum of !)
= 180° – ∠ADI – ∠AID
= ∠DAI (∠ sum of !)
= 90°
∠BHG = 180° – 90° (adj. ∠s on st. line)
= 90°
III is true.
The answer is D.

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HKDSE Exam Series — Master 1, Tackle 5 Hot Question Types for Mathematics (Compulsory Part) (New Syllabus Edition) Solution Guide

Instant Practice 15
1. Join OC.
OB = OC (radii)
` ∠OCB = ∠OBC (base ∠s, isos. !)
= 76°
∠BOC + ∠OBC + ∠OCB = 180° (∠ sum of !)
` ∠BOC = 180° – 76° – 76°
= 28°
ce
∠BOC = 2∠BAC (∠ at centre twice ∠ at )

` ∠BAC = (28°)

= 14°
The answer is A.

2. ∠BCA = ∠CAD (alt. ∠s, BC // AD)

= 28°
∠ACD = 90° (∠ in semi-circle)
∠BAD + ∠BCD = 180° (opp. ∠s, cyclic quad.)
` ∠CAB + 28° + 28° + 90° = 180°
∠CAB = 34°
The answer is C.

3. Join CD.
∠BCA = ∠CAD (alt. ∠s, BC // AD)
∠ACD = 90° (∠ in semi-circle)
∠BAD + ∠BCD = 180° (opp. ∠s, cyclic quad.)
20° + ∠CAD + ∠BCA + 90° = 180°
2∠CAD = 70°
∠CAD = 35°
ce
∠COD = 2∠CAD (∠ at centre twice ∠ at )
= 2(35°)
= 70°
The answer is D.

4. Join QT.
∠TQS = 90° (∠ in semi-circle)
∠RQS = ∠RPS + ∠QST (ext. ∠ of !)
= 24° + 14°
= 38°
∠RQT + ∠RST = 180° (opp. ∠s, cyclic quad.)
90° + 38° + ∠RSQ + 14° = 180°
∠RSQ = 38°
The answer is D.

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 12 Basic Geometry

5. Join QS and QT.

ce
∠QOT = 2∠QST (∠ at centre twice ∠ at )

∠QST = (20°)

= 10°
∠TQS = 90° (∠ in semi-circle)
∠RQS = ∠RPS + ∠QST (ext. ∠ of !)
= 24° + 10°
= 34°
∠RQT + ∠RSP = 180° (opp. ∠s, cyclic quad.)
90 + 34° + ∠RSP = 180°
°
∠RSP = 56°
The answer is C.

Instant Practice 16
1. Join OB.
ce
Reflex ∠BOD = 2∠BAD (∠ at centre twice ∠ at )
= 2(100°)
= 200°
OC = OB (radii)
∠OBC = ∠OCB (base ∠s, isos. !)
= 30°
∠BOC + ∠OBC + ∠OCB = 180° (∠ sum of !)
∠BOC = 180° – 30° – 30°
= 120°
` ∠COD = 200° – 120°
= 80°
The answer is B.

2. Join ED.
Reflex ∠AOD + ∠AOD = 360° (∠s at a pt.)
Reflex ∠AOD = 360° – 150°
= 210°
ce
Reflex ∠AOD = 2∠AED (∠ at centre twice ∠ at )

∠AED = (210°)

= 105°
∠BED + ∠BCD = 180° (opp. ∠s, cyclic quad.)
∠BED = 180° – 115°
= 65°
∠AEB = 105° – 65°
= 40°
The answer is C.

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HKDSE Exam Series — Master 1, Tackle 5 Hot Question Types for Mathematics (Compulsory Part) (New Syllabus Edition) Solution Guide

3. Join ED.
∠EDB + ∠EAB = 180° (opp. sides, cyclic quad.)
∠EDB = 180° – 100°
= 80°
∠EDC = 80° + 20°
= 100°
ce
Reflex ∠EOC = 2∠EDC (∠ at centre twice ∠ at )
= 2(100°)
= 200°
Reflex ∠EOC + ∠EOC = 360° (∠s at a pt.)
∠EOC = 360° – 200°
= 160°
The answer is D.

4. Join CD.
∠CAD = ∠CED (∠s in the same segment)
= 42°
∠ACD = 90° (∠ in semi-circle)
∠CAD + ∠ACD + ∠CDA = 180° (∠ sum of !)
∠CDA = 180° – 90° – 42°
= 48°
∠ABC + ∠CDA = 180° (opp. ∠s, cyclic quad.)
∠ABC = 180° – 48°
= 132°
The answer is C.

5. ∠BAD = 25° + 39°

= 64°
BA = BD (given)
` ∠ADB = ∠BAD (base ∠s, isos. !)
= 64°
∠ACB = ∠ADB (∠s in the same segment)
= 64°
∠AEB + ∠DAC = ∠ACB (ext. ∠ of !)
∠AEB = 64° – 25°
= 39°
The answer is B.

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