Mathematics in Focus Book 2B Chapter 10
Question Bank
Chapter 10 Deductive Geometry (I)
Warm-up
Conventional — level 1
2B10WC1_001E
Find the value of x in the figure.
40
x
23
Solution
( sum of △)
x + 23 + 40 = 180
x + 63 = 180
x = 117
Answer
117
2B10WC1_002E
Find the values of x and y in the figure.
98
x
57
y
Solution
x + 57 = 98
(ext.  of △)
x = 41
y + 57 = 180
(adj.  on st. line)
y = 123
Answer
x = 41, y = 123
2B10WC1_003E
In the figure, BCD is a straight line. AE // BD, CA // DE and BA = CA. Find the values of x, y and z.
A
E
y
65
B
z
x
D
C
1
© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
Question Bank
Solution
∵ BA = CA
(given)
∴ x = 65
(base s isos. △)
y=x
(alt. s, AE // BD)
= 65
(int. s, CA // DE)
y + z = 180
65 + z = 180
z = 115
Answer
x = 65, y = 65, z = 115
2B10WC1_004E
Determine which two triangles are necessarily congruent. Give reasons.
H
A
Q
C
R
B
G
P
I
Solution
△ABC  △PQR (SAS)
Answer
△ABC  △PQR (SAS)
2B10WC1_005E
Are △ABC and △XYZ similar? If they are, give reasons.
X
B
Y
C
A
Z
Solution
Yes, △ABC ~ △XYZ. (AAA)
Answer
yes, △ABC ~ △XYZ (AAA)
2B10WC1_006E
Are △ABC and △DEC similar? If they are, give reasons.
B
8
C
4
A
6
D
4.5
3
6
E
2
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Mathematics in Focus Book 2B Chapter 10
Question Bank
Solution
Yes, △ABC ~ △DEC. (3 sides proportional)
Answer
yes, △ABC ~ △DEC (3 sides proportional)
2B10WC1_007E
In the figure, AB = CD, AD = CB, BAD = 60 and ABD = 75.
B
C
x
(a) Name a pair of congruent triangles. Give reasons.
75
(b) Find x.
60
A
Solution
D
(a) △ABD  △CDB (SSS)
(b) In △ABD,
BDA + 60 + 75 = 180
( sum of △)
BDA + 135 = 180
BDA = 45
∵ △ABD  △CDB
∴ DBC = BDA
(corr. s, △s)
x = 45
Answer
(a) △ABD  △CDB (SSS)
(b) 45
2B10WC1_008E
In the figure, ABC and AED are straight lines. AB = BC = 3,
A
AE = ED = 4 and BE = 3.5.
3
(a) Name a pair of similar triangles. Give reasons.
B
(b) Find CD.
3
Solution
(a) △ABE ~ △ACD (ratio of 2 sides, inc. )
C
4
3.5
E
4
D
(b) ∵ △ABE ~ △ACD
∴
CD
AC
=
BE
AB
CD
33
=
3.5
3
CD =
6
 3.5
3
=7
Answer
(a) △ABE ~ △ACD (ratio of 2 sides, inc. ) (b) 7
3
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Mathematics in Focus Book 2B Chapter 10
Question Bank
Section 10.1
Conventional — level 1
2B101C1_001E
In the figure, prove that ABC is a straight line.
D
141
39
A
C
B
Solution
ABD + CBD = 141 + 39
= 180
(adj. s supp.)
∴ ABC is a straight line.
Answer
no numerical answer
2B101C1_002E
In the figure, prove that ABC is a straight line.
D
E
40
50
A
C
B
Solution
ABD + DBE + EBC = 90 + 40 + 50
= 180
(adj. s supp.)
∴ ABC is a straight line.
Answer
no numerical answer
2B101C1_003E
In the figure, prove that ABC is a straight line.
E
C
D
55
82
43
B
A
4
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Mathematics in Focus Book 2B Chapter 10
Question Bank
Solution
ABD + DBE + EBC = 43 + 82 + 55
= 180
(adj. s supp.)
∴ ABC is a straight line.
Answer
no numerical answer
2B101C1_004E
In the figure, prove that ABC is a straight line.
E
D
A
140
20
20
C
B
Solution
ABD + DBE + EBC = 20 + 20 + 140
= 180
(adj. s supp.)
∴ ABC is a straight line.
Answer
no numerical answer
2B101C1_005E
In the figure, ABD = a + 10, DBE = 2a, EBF = 85 – a and FBC = 85 – 2a. Prove that ABC
is a straight line.
E
D
A
F
2a
85 – a
85 – 2a
a +10
C
B
Solution
ABD + DBE + EBF + FBC = (a + 10) + 2a + (85 – a) + (85 – 2a)
= (a + 2a – a – 2a) + (10 + 85 + 85)
= 180
∴ ABC is a straight line.
(adj. s supp.)
Answer
no numerical answer
5
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Mathematics in Focus Book 2B Chapter 10
Question Bank
2B101C1_006E
In the figure, AFB = x, BFC = 30 – y, CFD = 70 + y and DFE = 80 – x. Prove that AFE is
a straight line.
C
B
30 – y
D
70 + y
80 – x
x
A
E
F
Solution
AFB + BFC + CFD + DFE = x + (30 – y) + (70 + y) + (80 – x)
= (x – y + y – x) + (30 + 70 + 80)
= 180
∴ AFE is a straight line.
(adj. s supp.)
Answer
no numerical answer
2B101C1_007E
In the figure, CED is a straight line. AEC = 44, AED = 4x and DEB = x + 10. Prove that
AEB is a straight line.
D
A
4x
x + 10
44
E
C
B
Solution
CEA + AED = 180
(adj. s on st. line)
44 + 4x = 180
4x = 136
x = 34
AED + BED = 4x + (x + 10)
= 5x + 10
= 5(34) + 10
= 180
∴ AEB is a straight line.
(adj. s supp.)
Answer
no numerical answer
6
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Mathematics in Focus Book 2B Chapter 10
Question Bank
2B101C1_008E
In the figure, AFB, CGD and EFGH are straight lines.
E
EFB = 70 and HGD = 110. Prove that AB // CD.
HGD + DGF = 180
B
G
C
Solution
70
F
A
D
110
(adj. s on st. line)
110 + DGF = 180
H
DGF = 70
∵ DGF = BFE
(corr. s equal)
∴ AB // CD
Answer
no numerical answer
2B101C1_009E
In the figure, AFB, CGD and EFGH are straight lines.
E
BFG = 58 and CGH = 122. Prove that AB // CD.
F
A
B
58
G
C
Solution
FGD = 122
D
122
(vert. opp. s)
H
BFG + FGD = 58 + 122
= 180
(int. s supp.)
∴ AB // CD
Answer
no numerical answer
2B101C1_010E
In the figure, AFB, CGD and EFGH are straight lines. BFG = 131 – k
E
and DGF = 49 + k. Prove that AB // CD.
F
A
131 – k
Solution
G
C
BFG + FGD = (131 – k) + (49 + k)
49 + k
B
D
= 131 + 49 – k + k
H
= 180
∴ AB // CD
(int. s supp.)
Answer
no numerical answer
7
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Mathematics in Focus Book 2B Chapter 10
Question Bank
2B101C1_011E
In the figure, ABCD is a straight line. ABE = FCD = a.
F
Prove that EB // CF.
B
A
a
C
D
a
Solution
EBD = 180 – a
(adj. s on st. line)
ACF = 180 – a
(adj. s on st. line)
E
∴ EBD = ACF
(alt. s equal)
∴ EB // CF
Answer
no numerical answer
2B101C1_012E
In the figure, EBF, GCH and ABCD are straight lines.
ABE = x, ABF = 5x and BCG = 90 – 2x.
(a) Find the value of x.
90 – 2x
x
A
(b) Prove that EF // GH.
G
E
C
B
H
F
Solution
(a) ABF + ABE = 180
D
5x
(adj. s on st. line)
5x + x = 180
6x = 180
x = 30
(b) ABE = 30
BCG = 90 – 2(30)
= 30
∵ ABE = BCG
∴ EF // GH
(corr. s equal)
Answer
(a) x = 30
(b) no numerical answer
2B101C1_013E
In the figure, EBF, GCH and ABCD are straight lines.
E
G
EBC = 7m, FBC = 3m, BCG = 54. Prove that EF // GH.
A
7m
B
54
3m
C
D
F
8
H
© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
Question Bank
Solution
EBC + FBC = 180
(adj. s on st. line)
7m + 3m = 180
10m = 180
m = 18
CBF = 3(18)
= 54
∵ CBF = BCG
(alt. s equal)
∴ EF // GH
Answer
no numerical answer
2B101C1_014E
In the figure, ADC, AEF and BDE are straight lines. ABD = u,
B
DAE = v and BEF = w. Prove that BAD = w – u – v.
C
u
D
w
v
Solution
A
In △ABE,
BAE + ABE = BEF
F
E
(ext.  of △)
(BAD + v) + u = w
BAD = w – u – v
Answer
no numerical answer
2B101C1_015E
In the figure, BCD and ACEF are straight lines.BAC = a,
ABC = b, FED = c and CDE = d.
F
E
B
(a) Prove that ACB = c – d.
c
b
C
(b) Prove that a + b + c = 180 + d.
a
d
D
Solution
A
(a) In △CDE,
DCE + CDE = FED
(ext.  of △)
DCE + d = c
DCE = c – d
ACB = DCE
(vert. opp. s)
=c–d
9
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Mathematics in Focus Book 2B Chapter 10
Question Bank
(b) In △ABC,
ABC + BAC + ACB = 180
( sum of △)
b + a + (c – d) = 180
a + b + c = 180 + d
Answer
(a) no numerical answer
(b) no numerical answer
2B101C1_016E
In the figure, △BCD is a right-angled triangle. ABD = 5a,
A
CBD = 30 and BDC = 2a. Prove that ABC is a straight line.
5a
B
30
Solution
2a
In △BCD,
D
C
2a + 90 + 30 = 180
( sum of △)
2a = 60
a = 30
CBD + ABD = 30 + 5(30)
= 180
∴ ABC is a straight line.
(adj. s supp.)
Answer
no numerical answer
2B101C1_017E (TSA)
In the figure, ABC = 50, ACB = 70 and ACD = 60.
A
D
Prove that AB // DC.
60
50
Solution
B
In △ABC,
BAC + 70 + 50 = 180
70
C
( sum of △)
BAC = 60
∵ BAC = ACD
∴ AB // DC
(alt. s equal)
Answer
no numerical answer
10
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Mathematics in Focus Book 2B Chapter 10
Question Bank
2B101C1_018E
In the figure, ABCD is a straight line. ABE = x, DCE = 2y
D
and BEC = 3x – 180. Prove that x = y.
C
B
2y
x
A
3x – 180
E
Solution
ABE + CBE = 180
(adj. s on st. line)
CBE = 180 – x
In △BCE,
CBE + BEC = DCE
(ext.  of △)
(180 – x) + (3x – 180) = 2y
2x = 2y
x=y
Answer
no numerical answer
11
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Mathematics in Focus Book 2B Chapter 10
Question Bank
Conventional — level 2
2B101C2_001E
In the figure, ABC is a straight line. ABD = DBE, ABF = 135
E
and CBE = 90. Prove that DBF is a straight line.
D
A
C
B
135
Solution
F
ABD = DBE
(given)
ABD + DBE + 90 = 180
(adj. s on st. line)
2ABD = 90
ABD = 45
ABD + ABF = 45 + 135
= 180
(adj. s supp.)
∴ DBF is a straight line.
Answer
no numerical answer
2B101C2_002E
In the figure, AFD is a straight line. AFB = 35,
EFD = 2a + 15, DFC = a and BFC = 90.
(a) Find the value of a.
A
E
B
(b) Prove that EFC is a straight line.
F
2a + 15
35
a
D
C
Solution
(a) 35 + 90 + a = 180
(adj. s on st. line)
a = 55
(b) EFD + CFD = [2(55) + 15] + (55)
= 180
∴ EFC is a straight line.
(adj. s supp.)
Answer
(a) a = 55
(b) no numerical answer
12
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Mathematics in Focus Book 2B Chapter 10
Question Bank
2B101C2_003E
In the figure, DBI, EBH and FBG are straight lines.
E
F
ABD = DBE = 2x, EBF = x, FBC = 3x and CBI = IBH.
(a) Find the value of x.
x
D
2x
(b) Prove that ABC is a straight line.
3x
C
2x
B
A
I
H
G
Solution
(a) IBH = DBE = 2x
(vert. opp. s)
CBI = IBH
(given)
= 2x
DBE + EBF + FBC + CBE = 180
(adj. s on st. line)
2x + x + 3x + 2x = 180
8x = 180
x = 22.5
(b) ABD + DBE + EBF + FBC = 2(22.5) + 2(22.5) + (22.5) + 3(22.5)
= 180
(adj. s supp.)
∴ ABC is a straight line.
Answer
(a) 22.5
(b) no numerical answer
2B101C2_004E (TSA)
In the figure, DEF is a straight line. ABC = 116,
D
A
CBE = BEF = 122.
(a) Find ABE.
116
(b) Prove that AB // DF.
B
122
E
122
F
C
Solution
(a) ABE + ABC + CBE = 360
(s at a pt.)
ABE + 116 + 122 = 360
ABE = 122
(b) ∵ ABE = BEF = 122
∴ AB // DF
(alt. s equal)
Answer
(a) 122
(b) no numerical answer
13
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Mathematics in Focus Book 2B Chapter 10
Question Bank
2B101C2_005E (TSA)
In the figure, FA // EB, FAB = 80, ABC = 125
and BCD = 45.
C
A
45
125
80
(a) Find ABE and EBC.
B
(b) Prove that EB // DC.
D
F
Solution
E
(a) ABE + 80 = 180
(int. s, FA // EB)
ABE = 100
EBC + ABE + ABC = 360
(s at a pt.)
EBC + 100 + 125 = 360
EBC = 135
(b) EBC + BCD = 135 + 45
= 180
(int. s supp.)
∴ EB // DC
Answer
(a) ABE = 100, EBC = 135
(b) no numerical answer
2B101C2_006E
In the figure, ABC, CEG and DEF are straight lines. AC // DF.
ABD = 2m + 8, BDE = 3m – 12 and GEF = 132.
B
A
C
2m + 8
(a) Find the value of m.
(b) Prove that BD // CG.
3m – 12
D
Solution
(a)
ABD = BDE
(alt. s, AC // DF)
E
F
132
G
2m + 8 = 3m – 12
m = 20
(b) BDE = 3(20) – 12
= 48
CEF + 132 = 180
(adj. s on st. line)
CEF = 48
∵ BDE = CEF
∴ BD // CG
(corr. s equal)
Answer
(a) 20
(b) no numerical answer
14
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Mathematics in Focus Book 2B Chapter 10
Question Bank
2B101C2_007E
In the figure, AFB, CGD and EFGH are straight lines.
H
C
IF // JG. EFI = 55, AFG = 135 and JGD = 80.
(a) Find CGF.
A
G
80
135
(b) Prove that AB // CD.
F
55
D
J
B
E
I
Solution
(a) FGJ = EFI = 55
(corr. s, IF // JG)
CGF + FGJ + JGD = 180
(adj. s on st. line)
CGF + 55 + 80 = 180
CGF = 45
(b) AFG + CGF = 135 + 45
= 180
(int. s supp.)
∴ AB // CD
Answer
(a) 45
(b) no numerical answer
2B101C2_008E
In the figure, DAB = 41, BCE = 77 and
A
D
41
reflex  ABC = 242. Prove that AD // CE.
242
B
77
E
C
Solution
Add a line FBG such that AD // FG.
ABG + DAB = 180
(int. s, AD // FG)
ABG + 41 = 180
A
ABG = 139
D
41
CBG = 242 – ABG
F
= 242 – 139
B
242
G
= 103
77
CBG + BCE = 103 + 77
C
E
= 180
∴ FG // CE
(int. s supp.)
∴ AD // CE
Answer
no numerical answer
15
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Mathematics in Focus Book 2B Chapter 10
Question Bank
2B101C2_009E
In the figure, ABC = 25, CDE = 130 and
D
reflex  DCB = 285. Prove that AB // DE.
E
130
C
285
Solution
25
A
B
Add a line CF such that DE // CF.
DCF + CDE = 180
(int. s, DE // CF)
D
DCF + 130 = 180
E
130
DCF = 50
BCF + DCF + reflex  DCB = 360
(s at a pt.)
285
BCF + 50 + 285 = 360
F
25
A
BCF = 25
C
B
∵ ABC = BCF
(alt. s equal)
∴ AB // CF
∴ AB // DE
Answer
no numerical answer
2B101C2_010E
In the figure, ABCD, EGC and FGB are straight lines.
F
AEC = EAC = x, BDF = DFB = y and EGF = z.
E
Prove that 2x + 2y – z = 180.
x
y
z
G
Solution
y
x
In △AEC,
A
ECA + x + x = 180
B
C
D
( sum of △)
ECA = 180 – 2x
In △BFD,
FBD + y + y = 180
( sum of △)
FBD = 180 – 2y
BGC = z
(vert. opp. s)
In △BGC,
GCB + GBC + BGC = 180
(180 – 2x) + (180 – 2y) + z = 180
( sum of △)
360 – 2x – 2y + z = 180
2x + 2y – z = 180
Answer
no numerical answer
16
© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
Question Bank
Conventional — level 3
2B101C3_001E (+2B08)
In the figure, ABCD, AKFJ, DLGJ, EKBI, ICLH and EFGH
I
are straight lines. BCLGFK is a regular hexagon.
B
A
(a) Find the size of each interior angle of the hexagon BCLGFK.
(b) Prove that AD // EH.
C
L
K
E
Solution
D
G
F
H
J
Suggested marks: 5 marks
(a) Sum of interior angles of the hexagon BCLGFK
= (6 – 2) × 180
( sum of polygon)
(1M)
= 720
Size of each interior angle of the hexagon BCLGFK
=
720
6
= 120
(1A)
(b) ABK + KBC = 180
(adj. s on st. line)
ABK + 120 = 180
ABK = 60
AKB + BKF = 180
(adj. s on st. line)
AKB + 120 = 180
AKB = 60
In △ABK,
BAK + ABK + AKB = 180
( sum of △)
(1M)
BAK + 60 + 60 = 180
BAK = 60
BAK + KFG = 60 + 120
(1M)
= 180
∴ AD // EH
(int. s supp.)
(1A)
Answer
(a) 120
(b) no numerical answer
17
© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
Question Bank
2B101C3_002E
In the figure, ABCD and EFGH are straight lines.
B
A
FB // GC and BI // CJ.
D
C
△BIF and △CJG are two right-angled triangles.
J
I
(a) Prove FBI = GCJ.
(b) Prove BFI = CGJ.
E
(c) Prove that FI // GJ.
F
G
H
Solution
Suggested marks: 10 marks
(a) ABF = BCG
(corr. s, FB // GC)
(1M)
IBC = JCD
(corr. s, BI // CJ)
(1M)
FBI + ABF + IBC = 180
(adj. s on st. line)
(1M)
(adj. s on st. line)
(1M)
FBI = 180 – ABF – IBC
GCJ + BCG + JCD = 180
GCJ + ABF + IBC = 180
GCJ = 180 – ABF – IBC
∴ FBI = GCJ
(1A)
(b) In △BFI,
BFI + BIF + FBI = 180
( sum of △)
(1M)
( sum of △)
(1M)
BFI + 90 + FBI = 180
BFI = 90 – FBI
In △CGJ,
CGJ + CJG + GCJ = 180
CGJ + 90 + GCJ = 180
CGJ = 90 – GCJ
∵ FBI = GCJ
(proved in (a))
∴ BFI = CGJ
(1M)
(c) BFE = CGF
(corr. s, FB // GC)
(1M)
(corr. s equal)
(1A)
IFE = BFI + BFE
JGF = CGJ + CGF
= BFI + BFE
∴ IFE = JGF
∴ FI // GJ
Answer
(a) no numerical answer
(b) no numerical answer
(c) no numerical answer
18
© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
Question Bank
Multiple Choice Questions — level 1
2B101M1_001E
Refer to the figure. Which of the following must be true?
I.
AOB is a straight line.
II.
COD is a straight line.
D
E
60
III. EOF is a straight line.
A
A. I and II only
B.
C.
D. I, II and III
II and III only
75
50
B
70 O 45
60
I and III only
F
C
Solution
For I:
AOE + EOD + DOB = 50 + 60 + 75
= 185
≠180
∴ AOB is not a straight line.
i.e. I cannot be true.
For II:
COF + FOB + BOD = 60 + 45 + 75
= 180
∴ COD is a straight line.
(adj. s supp.)
i.e. II must be true.
For III:
EOD + DOB + BOF = 60 + 75 + 45
= 180
∴ EOF is a straight line.
(adj. s supp.)
i.e. III must be true.
∴ The answer is C.
Answer
C
2B101M1_002E
In the figure, AEGB and CFHD are straight lines.
I.
AC // BD
II.
EF // GH
G
B
125
70
55
C
III. AB // CD
A. I and II only
B.
C.
D. I, II and III
II and III only
E
A
Which of the following must be true?
F
110
H
125
D
I and III only
19
© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
Question Bank
Solution
For I:
ACD + CDB = 55 + 125
= 180
(int. s supp.)
∴ AC // BD
i.e. I must be true.
For II:
EFH + GHF = 70 + 110
= 180
(int. s supp.)
∴ EF // GH
i.e. II must be true.
For III:
CAB + ACD = 125 + 55
= 180
(int. s supp.)
∴ AB // CD
i.e. III must be true.
∴ The answer is D.
Answer
D
2B101M1_003E
Refer to the figure. Which of the following must be true?
I.
△ABC is an obtuse-angled triangle.
II.
ACD is a straight line.
B
25
III. AB is parallel to ED.
A. I and II only
B.
C.
D. I, II and III
II and III only
I and III only
Solution
A
67
42
C
43
D
E
For I:
In △ABC,
BCA + 25 + 42 = 180
( sum of △)
BCA = 113
∴ △ABC is an obtuse-angled triangle.
i.e. I must be true.
For II:
ACB + BCD = 113 + 67
= 180
∴ ACD is a straight line.
(adj. s supp.)
i.e. II must be true.
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Mathematics in Focus Book 2B Chapter 10
Question Bank
For III:
∵ BAD ≠ ADE
∴ AB is not parallel to ED.
i.e. III cannot be true.
∴ The answer is A.
Answer
A
2B101M1_004E
In the figure, BCD = BDC. Which of the following must be true?
I.
BCD = 70
II.
ABC is a straight line.
A
B
III. CDE is a straight line.
40
A. I and II only
B.
C.
D. I, II and III
II and III only
140
I and III only
Solution
110
C
D
E
For I:
BCD = BDC
(given)
In △BCD,
BCD + BDC + CBD = 180
( sum of △)
2BCD + 40 = 180
2BCD = 140
BCD = 70
i.e. I must be true.
For II:
ABD + CBD = 140 + 40
= 180
∴ ABC is a straight line.
(adj. s supp.)
i.e. II must be true.
For III:
BDC = BCD = 70
BDC + BDE = 70 + 110
= 180
∴ CDE is a straight line.
(adj. s supp.)
i.e. III must be true.
∴ The answer is D.
Answer
D
21
© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
Question Bank
Multiple Choice Questions — level 2
2B101M2_001E
In the figure, ABC, DEFG, HBE and CFI are straight lines.
H
If a = b, which of the following must be true?
I.
AC // DG
II.
c=d
a
B
A
C
c
b
III. HE // CI
D
A. I and II only
d
E
G
F
B. I and III only
I
C. II and III only
D. I, II and III
Solution
For I:
a=b
(given)
∴ AC // DG
(corr. s equal)
i.e. I must be true.
For II:
(alt. s, AC // DG)
c=d
i.e. II must be true.
For III:
Consider the case that a = b = 120 and c = d = 50.
ABE + ABH = 180
(adj. s on st. line)
ABE + 120 = 180

ABE = 60
∵ ABE ≠ ACI
∴ HE is not parallel to CI in this case.
∴ HE may not be parallel to CI.
i.e. III may not be true.
∴ The answer is A.
Answer
A
2B101M2_002E
In the figure, BCD is a straight line. BA // CE and AE // BC.
E
A
Which of the following must be true?
e
a
A. c + d = f
B. b + e = 180
b
B
C. a = c
f
c
d
C
D
D. a + e + f = 180
22
© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
Question Bank
Solution
For choice A:
In △CDE,
(ext.  of △)
f+d=c
∴ Choice A is not true.
For choice B:
Consider the case that a = 60, b = 120, c = 60, e = 120, f = 30 and d = 30.
b + e = 120 + 120 ≠ 180
∴ b + e ≠ 180in this case.
∴ Choice B may not be true.
For choice C:
(int. s, AE // BC)
a + b = 180
b = 180 – a
b + c = 180
(int. s, BA // CE)
(180 – a) + c = 180
–a+c=0
a=c
∴ Choice C must be true.
For choice D:
a + e = 180
(int. s, BA // CE)
Since f ≠ 0, a + e + f ≠ 180.
∴ Choice D is not true.
∴ The answer is C.
Answer
C
2B101M2_003E (TSA)
In the figure, AGB, CHD, EGC and FGHI are straight lines.
F
E
Prove that AB // CD. Which of the following proofs is incorrect?
A. BGH = 110
(vert. opp. s)
110
A
CGB = CGH + BGH
C
= 140
CGB + GCH = 180
∴ AB // CD
B.
G
30
40
H
B
D
I
(int. s supp.)
FGA + AGC + CGH = 180
(adj. s on st. line)
AGC = 40
GCH = AGC
∴ AB // CD
(alt. s equal)
23
© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
C.
AGC = 40
Question Bank
(alt. s, AB // CD)
AGH = 40
GHD = 30
(ext.  of △)
AGH = GHD
∴ AB // CD
D. CHG + 30 + 40 = 180
(alt. s equal)
( sum of △)
CHG = 110
AGF = CHG
∴ AB // CD
(corr. s equal)
Solution
For choice C:
Since we have not proved that AB // CD, we cannot use the theorem ‘alt. s, AB // CD’.
∴ The proof in choice C is incorrect.
∴ The answer is C.
Answer
C
24
© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
Question Bank
Section 10.2
Conventional — level 1
2B102C1_001E (TSA)
Refer to the following figure. Prove that △ABC  △ADC.
A
B
D
Solution
AB = AD
(given)
BAC = DAC
(given)
AC = AC
(common side)
∴ △ABC  △ADC
(SAS)
C
Answer
no numerical answer
2B102C1_002E (TSA)
Refer to the following figure. Prove that △ABC  △ADC.
B
C
A
Solution
AB = AD
(given)
BC = DC
(given)
AC = AC
(common side)
∴ △ABC  △ADC
(SSS)
D
Answer
no numerical answer
2B102C1_003E (TSA)
In the figure, ABC = DBC and ACB = DCB.
Prove that △ABC  △DBC.
D
A
C
Solution
ABC = DBC
(given)
ACB = DCB
(given)
CB = CB
(common side)
∴ △ABC  △DBC
(ASA)
B
Answer
no numerical answer
25
© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
Question Bank
2B102C1_004E (TSA)
In the figure, ABC = ADC and ACB = ACD.
A
Prove that △ABC  △ADC.
C
B
Solution
ABC = ADC
(given)
ACB = ACD
(given)
AC = AC
(common side)
∴ △ABC  △ADC
(AAS)
D
Answer
no numerical answer
2B102C1_005E
In the figure, BDC is a straight line. BAD = CAD
A
and ACD = 90. Prove that △ABD  △ACD.
B
Solution
ADB + 90 = 180
C
D
(adj. s on st. line)
ADB = 90
∴ ADB = ADC
BAD = CAD
(given)
AD = AD
(common side)
∴ △ABD  △ACD
(ASA)
Answer
no numerical answer
2B102C1_006E
In the figure, ACE and BCD are straight lines.
A
CB = CD and AB // DE. Prove that △ABC  △EDC.
B
C
Solution
D
BAC = DEC
(alt. s, AB // DE)
ABC = EDC
(alt. s, AB // DE)
CB = CD
(given)
∴ △ABC  △EDC
(AAS)
E
Answer
no numerical answer
26
© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
Question Bank
2B102C1_007E
E
In the figure, ABC and EDB are straight lines. AD = EC,
DB = CB and ABD = 90. Prove that △ABD  △EBC.
D
Solution
A
DB = CB
(given)
90 + CBE = 180
(adj. s on st. line)
C
B
CBE = 90
∴ ABD = CBE = 90
AD = EC
(given)
∴ △ABD  △EBC
(RHS)
Answer
no numerical answer
2B102C1_008E
In the figure, AC intersects BD at E. ABE = DCE
D
and ACB = DBC. Prove that △ABC  △DCB.
C
E
Solution
ACB = DBC
(given)
BC = CB
(common side)
ABE = DCE
(given)
A
B
ABC = ABE + DBC
DCB = DCE + ACB
= ABE + DBC
∴ ABC = DCB
∴ △ABC  △DCB
(ASA)
Answer
no numerical answer
2B102C1_009E
In the figure, ABC is a straight line. DAB = 64
D
and DBC = 122. Prove that AB = AD.
122
64
A
27
B
C
© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
Question Bank
Solution
In △ABD,
ABD + 122 = 180
(adj. s on st. line)
ABD = 58
ADB + 64 = 122
(ext.  of △)
ADB = 58
∴ ABD = ADB
(sides opp. equal s)
∴ AB = AD
Answer
no numerical answer
2B102C1_010E
In the figure, ABC is a straight line. ADC = 110, DBC = 80
D
and BCD = 30. Prove that △ABD is an isosceles triangle.
80
A
110
30
C
B
Solution
In △BDC,
BDC + 80 + 30 = 180
( sum of △)
BDC = 70
ADB = ADC – BDC
= 110 – 70
= 40
In △ADC,
DAC + 110 + 30 = 180
( sum of △)
DAC = 40
∴ ADB = DAB
∴ AB = BD
(sides opp. equal s)
i.e. △ABD is an isosceles triangle.
Answer
no numerical answer
2B102C1_011E
In the figure, ADC is a straight line. BAD = 66, ADB = 90
A
and DBC = 33. Prove that △ABC is an isosceles triangle.
66
D
B
28
33
C
© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
Question Bank
Solution
In △ABD,
ABD + 90 + 66 = 180
( sum of △)
ABD = 24
ABC = ABD + DBC
= 24 + 33
= 57
In △BDC,
DCB + 33 = 90
(ext.  of △)
DCB = 57
∴ ABC = ACB
(sides opp. equal s)
∴ AB = AC
i.e. △ABC is an isosceles triangle.
Answer
no numerical answer
2B102C1_012E
In the figure, △ABD is an isosceles triangle with AD = BD.
C
BAD = BDC. Prove that AB // DC.
B
A
Solution
ABD = BAD
(base s, isos. △)
BAD = BDC
(given)
D
∴ ABD = BDC
∴ AB // DC
(alt. s equal)
Answer
no numerical answer
29
© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
Question Bank
Conventional — level 2
2B102C2_001E
In the figure, BA // ED, BC // EF and CA // FD. D is a point on BC
A
and BC = EF.
(a) Prove that ABC = DEF.
(b) Prove that △ABC  △DEF.
B
C
D
F
E
Solution
(a) BDE = DEF
(alt. s, BC // EF)
ABC = BDE
(alt. s, BA // ED)
∴ ABC = DEF
(b) BC = EF
(given)
ABC = DEF
(proved in (a))
ACB = CDF
(alt. s, CA // FD)
CDF = DFE
(alt. s, BC // EF)
∴ ACB = DFE
∴ △ABC  △DEF
(ASA)
Answer
(a) no numerical answer
(b) no numerical answer
2B102C2_002E
In the figure, AD and BE intersects at C. AC = DC and BC = EC.
A
B
(a) Prove that △ABC  △DEC.
(b) Prove that AB // ED.
C
E
D
Solution
(a) AC = DC
(given)
BC = EC
(given)
ACB = DCE
(vert. opp. s)
∴ △ABC  △DEC
(SAS)
(b) ∵ △ABC  △DEC
(proved in (a))
∴ ABC = DEC
(corr. s, △s)
∴ AB // ED
(alt. s equal)
Answer
(a) no numerical answer
(b) no numerical answer
30
© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
Question Bank
2B102C2_003E
In the figure, BGHE is a straight line. BA // DE,
A
BC // FE, BA = DE and BC = FE.
F
(a) Prove that △ABC  △DEF.
G
B
(b) Prove that AC // FD.
H
E
C
D
Solution
(a) BA = ED
(given)
BC = EF
(given)
ABE = DEB
(alt. s, BA // DE)
CBE = FEB
(alt. s, BC // FE)
ABC = ABE + CBE
DEF = DEB + FEB
= ABE + CBE
∴ ABC = DEF
∴ △ABC  △DEF
(b) △ABC  △DEF
(SAS)
(proved in (a))
∴BAC = EDF
(corr. s, △s)
ABE = DEB
(proved in (a))
AGH = ABE + BAC
(ext.  of △)
DHG = DEB + EDF
(ext.  of △)
= ABE + BAC
∴ AGH = DHG
∴ AC // FD
(alt. s equal)
Answer
(a) no numerical answer
(b) no numerical answer
2B102C2_004E
In the figure, ABCDEF is a regular hexagon and CA // DF.
A
F
(a) Prove that △ABC  △FED.
(b) Using the result of (a), prove that △ACG  △DFG.
G
B
C
E
D
Solution
(a) ∵ ABCDEF is a regular hexagon.
∴ AB = BC = CD = DE = EF = FA
ABC = FED
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Mathematics in Focus Book 2B Chapter 10
Question Bank
△ABC  △FED
(SAS)
(b) △ABC  △FED
(proved in (a))
CA = DF
(corr. sides, △s)
ACG = DFG
(alt. s, CA // DF)
CAG = FDG
(alt. s, CA // DF)
∴ △ABC  △FED
(ASA)
Answer
(a) no numerical answer
(b) no numerical answer
2B102C2_005E
In the figure, BCEF is a straight line. BA // CD,
A
D
AD // BF and BAD = DEF. Prove that △DCE
is an isosceles triangle.
B
C
E
F
Solution
Let BAD = DEF = x.
ADC + BAD = 180
(int. s, BA // CD)
ADC + x = 180
ADC = 180 – x
DCE = ADC
(alt. s, AD // BF)
= 180 – x
DEC + DEF = 180
(adj. s on st. line)
DEC + x = 180
DEC = 180 – x
∴ DCE = DEC
(sides opp. equal s)
∴ DC = DE
i.e. △DCE is an isosceles triangle.
Answer
no numerical answer
2B102C2_006E
In the figure, AC and BD intersect at E. AB = BC, BDC = 22.5 and ABC = ACD = 90. Prove
that △BCD is an isosceles triangle.
A
D
E
B
22.5
C
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Mathematics in Focus Book 2B Chapter 10
Question Bank
Solution
BAC = BCA
(base s, isos. △)
In △ABC,
BAC + ACB + 90 = 180
( sum of △)
2ACB = 90
ACB = 45
In △BCD,
DBC + BCD + BDC = 180
( sum of △)
DBC + (45 + 90) + 22.5 = 180
DBC = 22.5
∵ DBC = BDC
∴ BC = DC
(sides opp. equal s)
i.e. △BCD is an isosceles triangle.
Answer
no numerical answer
2B102C2_007E
In the figure, BEC is a straight line. AB = DC, BE = CE,
ABE = FEB = DCE = 90.
F
A
D
(a) Prove that AE = DE.
(b) Using the result of (a), prove that △AEF  △DEF.
B
E
C
Solution
(a) ABE = DCE = 90
(given)
AB = DC
(given)
BE = CE
(given)
∴ △ABE  △DCE
(SAS)
∴ AE = DE
(corr. sides, △s)
(b) AE = DE
AEB = DEC
(proved in (a))
(corr. s, △s)
AEF = 90 – AEB
FEC + 90 = 180
(adj. s on st. line)
FEC = 90
DEF = FEC – DEC
DEF = 90 – AEB
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Mathematics in Focus Book 2B Chapter 10
Question Bank
∴ AEF = DEF
FE = FE
(common side)
∴ △AEF  △DEF
(SAS)
Answer
(a) no numerical answer
(b) no numerical answer
2B102C2_008E
In the figure, ADB = ADC = 130 and DBC = DCB = 50.
A
Prove that △ABC is an isosceles triangle.
130
130
D
50
50
B
C
Solution
∵ DBC = DCB = 50
(given)
∴ BD = CD
(sides opp. equal s)
AD = AD
(common side)
ADB = ADC = 130
(given)
∴ △ABD  △ACD
(SAS)
∴ AB = AC
(corr. sides, △s)
i.e. △ABC is an isosceles triangle.
Answer
no numerical answer
2B102C2_009E
In the figure, BCDE is a straight line. AB = AE
A
and BC = DE. Prove ACD = ADC.
B
C
D
E
Solution
ABC = AED
(base s, isos. △)
AB = AE
(given)
BC = DE
(given)
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Mathematics in Focus Book 2B Chapter 10
Question Bank
∴ △ABC  △AED
(SAS)
∴ AC = AD
(corr. sides, △s)
∴ ACD = ADC
(base s, isos. △)
Alternative solution:
ABC = AED
(base s, isos. △)
AB = AE
(given)
BD = BC + CD
= DE + CD
= EC
∴ △ABD  △AEC
(SAS)
∴ ACD = ADC
(corr. s, △s)
Answer
no numerical answer
2B102C2_010E
In the figure, PS and TQ intersect at R. PQ = TS and QPT = STP.
Q
(a) Prove that △PQT  △TSP.
S
R
(b) Using the result of (a), prove that △PQR  △TSR.
(c) Prove that △PRT is an isosceles triangle.
P
T
Solution
(a) PQ = TS
(given)
QPT = STP
(given)
PT = TP
(common side)
∴ △PQT  △TSP
(SAS)
(b) △PQT  △TSP
(proved in (a))
∴ PQR = TSR
(corr. s, △s)
QRP = SRT
(vert. opp s)
PQ = TS
(given)
∴ △PQR  △TSR
(AAS)
(c) △PQR  △TSR
∴ PR = TR
(proved in (b))
(corr. sides, △s)
∴ △PRT is an isosceles triangle.
Answer
(a) no numerical answer
(b) no numerical answer
(c) no numerical answer
35
© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
Question Bank
2B102C2_011E
In the figure, NRQP is a straight line. OQ  NP and
MR  NP. NO = PM and NR = PQ.
O
N
(a) Prove that △NOQ  △PMR.
R
(b) Prove that △OPQ  △MNR.
Q
(c) Using the result of (a) and (b), prove that
M
P
NO // MP and MN // PO.
Solution
(a) OQN = MRP = 90
(given)
NO = PM
(given)
NR = PQ
(given)
NQ = NR + RQ
PR = PQ + RQ
= NR + RQ
∴ NQ = PR
∴ △NOQ  △PMR
(b) △NOQ  △PMR
(RHS)
(proved in (a))
∴ OQ = MR
(corr. sides, △s)
OQP = MRN = 90
(given)
PQ = NR
(given)
∴ △OPQ  △MNR
(SAS)
(c) △NOQ  △PMR
(proved in (a))
∴ ONQ = MPR
(corr. s, △s)
∴ NO // MP
(alt. s equal)
△OPQ  △MNR
(proved in (a))
∴ OPQ = MNR
(corr. s, △s)
∴ MN // PO
(alt. s equal)
Answer
(a) no numerical answer
(b) no numerical answer
(c) no numerical answer
36
© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
Question Bank
Conventional — level 3
2B102C3_001E
In the figure, ABD and ACE are straight lines. BE and CD intersect at F.
A
BD = CE and ABC = ACB.
(a) Prove that ADE = AED.
B
(b) Using the result of (a), prove that △BDE  △CED.
(c) Using the result of (b), prove that △DEF is an isosceles triangle.
C
F
D
E
Solution
Suggested marks: 6 marks
(a) ABC = ACB
(given)
AB = AC
(sides opp. equal s)
BD = CE
(given)
AD = AB + BD
(1M)
AE = AC + CE
= AB + BD
∴ AD = AE
∴ ADE = AED
(b) ADE = AED
(base s, isos. △)
(proved in (a))
BD = CE
(given)
DE = ED
(common side)
∴ △BDE  △CED
(SAS)
(c) △BDE  △CED
(1A)
(1M)
(1A)
(proved in (b))
BED = CDE
(corr. s, △s)
(1M)
∴ DF = EF
(sides opp. equal s)
(1A)
i.e. △DEF is an isosceles triangle.
Answer
(a) no numerical answer
(b) no numerical answer
(c) no numerical answer
37
© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
Question Bank
Multiple Choice Questions — level 1
2B102M1_001E
Refer to the figure. Which of the following reasons can be
C
D
used to prove that △ABC  △DCB?
A. SSS
B.
C.
D. RHS
SSA
SAS
Solution
A
AB = DC
(given)
BC = CB
(common side)
ABC = DCB = 90
(given)
∴ △ABC  △DCB
(SAS)
B
∴ The answer is B.
Answer
B
2B102M1_002E
In the figure, BDEC is a straight line. BAC = ADC = 90
A
and ABD = AED. Which of the following must be true?
I.
△ABD  △ACD
II.
△ACD  △BCA
B
III. △ABD  △AED
A. I only
B.
C.
D. II and III only
I and II only
D
E
C
III only
Solution
For I:
∵ △ABD and △ACD are not under SAS, ASA, AAS, SSS and RHS.
∴ △ABD may not be congruent to △ACD.
i.e. I may not be true.
For II:
∵ △ACD and △BCA are not under SAS, ASA, AAS, SSS and RHS.
∴ △ACD may not be congruent to △BCA.
i.e. II may not be true.
For III:
ABD = AED
(given)
ADB + 90 = 180
(adj. s on st. line)
ADB = 90
∴ ADB = ADE
AD = AD
(common side)
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Mathematics in Focus Book 2B Chapter 10
∴ △ABD  △AED
Question Bank
(AAS)
i.e. III must be true.
∴ The answer is B.
Answer
B
2B102M1_003E
In the figure, ADC is a straight line. BD  AC, AD = CD and BCD = 40.
Which of the following must be true?
I.
△ABD  △CBD
II.
DBC = 40
B
III. △ABC is an isosceles triangle.
40
A
A. I and II only
B.
I and III only
C.
II and III only
D
C
D. I, II and III
Solution
For I:
AD = CD
(given)
ADB = CDB = 90
(given)
BD = BD
(common side)
∴ △ABD  △CBD
(SAS)
i.e. I must be true.
For II:
In △BCD,
DBC + BCD = BDA
(ext.  of △)
DBC + 40 = 90
DBC = 50
i.e. II is not true.
For III:
∵ △ABD  △CBD
(proved)
∴ AB = CB
(corr. sides, △s)
i.e. △ABC is an isosceles triangle.
∴ III must be true.
∴ The answer is B.
Answer
B
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© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
Question Bank
2B102M1_004E
In the figure, ABCD is a straight line. AB = FB, FAB = 25°, BFC = 80
and ECD = 40. Which of the following must be true?
I.
FBC = 40
II.
△FBC is an isosceles triangle.
F
80
III. BF // CE
A
A. I only
B.
II only
C.
I and II only
E
40
25
B
C
D
D. II and III only
Solution
For I:
AFB = FAB
(base s, isos. △)
= 25
In △FAB,
FBC = FAB + AFB
(ext.  of △)
= 25 + 25
= 50
≠ 40
∴ I is not true.
For II:
In △FBC,
FCB + FBC + BFC = 180
( sum of △)
FCB + 50 + 80 = 180
FCB = 50
∵ FBC = FCB
∴ BF = CF
(sides opp. equal s)
i.e. △FBC is an isosceles triangle.
∴ II must be true.
For III:
∵ FBC ≠ ECD
∴ BF is not parallel to CE.
i.e. III is not true.
∴ The answer is B.
Answer
B
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© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
Question Bank
Multiple Choice Questions — level 2
2B102M2_001E
In the figure, DEFB is a straight line. Which of the following must be true?
I.
△ADE  △BCF
II.
ABD = CDB
A
B
F
III. AB // DC
E
A. I and II only
B.
I and III only
C.
II and III only
D
C
D. I, II and III
Solution
For I:
AD = CB
(given)
ADE = CBF
(given)
AED = CFB = 90
(given)
∴ △ADE  △CBF
(AAS)
i.e. I may not be true.
For II:
AD = CB
(given)
DB = BD
(common side)
ADB = CBD
(given)
∴ △ABD  △CDB
(SAS)
∴ ABD = CDB
(corr. s, △s)
i.e. II must be true.
For III:
∵ ABE = CDF
(proved)
∴ AB // DC
(alt. s equal)
i.e. III must be true.
∴ The answer is C.
Answer
C
2B102M2_002E
In the figure, ABCDE is a regular pentagon. AD and CE
A
intersect at F. Which of the following must be true?
I.
△ACD is an isosceles triangle.
II.
△DEF is an isosceles triangle.
E
B
F
III. △ACF is an isosceles triangle.
C
41
D
© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
Question Bank
A. I and II only
B.
I and III only
C.
II and III only
D. I, II and III
Solution
For I:
∵ ABCDE is a regular pentagon.
∴ AB = BC = CD = DE = AE and ABC = AED.
∴ △ABC  △AED
(SAS)
∴ AC = AD
(corr. sides, △s)
i.e. △ACD is an isosceles triangle.
i.e. I must be true.
For II:
Obviously, △AED  △CDE
(SAS)
∴ ADE = CED
(corr. s, △s)
∴ FD = FE
(sides opp. equal s)
i.e. △DEF is an isosceles triangle.
i.e. II must be true.
For III:
Obviously, △ABC  △CDE
(SAS)
∴ BAC = DCE
(corr. s, △s)
Similarly, △ABC  △DEA
(SAS)
∴ BCA = EAD
(corr. s, △s)
BAE = BCD
CAF = BAE – BAC – EAD
ACF = BCD – DCE – BCA
= BAE – BAC – EAD
∴ CAF = ACF
∴ AF = CF
(sides opp. equal s)
i.e. △ACF is an isosceles triangle.
i.e. III must be true.
∴ The answer is D.
Answer
D
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© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
Question Bank
2B102M2_003E (TSA)
In the figure, AB = CD and ABD = CDB. Prove that △ABD  △CDB.
Which of the following proofs is correct?
A. AB = CD
(given)
AD = CB
(given)
ABD = CDB
(given)
∴ △ABD  △CDB
(RHS)
AB = CD
(given)
BD = DB
(common side)
ABD = CDB
(given)
∴ △ABD  △CDB
(SAS)
AB = CD
(given)
BD = DB
(common side)
AD = CB
(given)
∴ △ABD  △CDB
(SSS)
B.
C.
D. AB = CD
D
A
B
C
(given)
ABD = CDB
(given)
ADB = CBD
(corr. s, △s)
∴ △ABD  △CDB
(AAS)
Solution
Since we do not know whether AD = CB before we prove that △ABD  △CDB, we cannot use
AD = CB to prove △ABD  △CDB.
∴ Choices A and C are not correct.
Since we have not proved that △ABD  △CDB, we cannot use the theorem ‘corr. s,  △s’.
∴ The proof in choice D is incorrect.
Only the proof in choice B is correct.
∴ The answer is B.
Answer
B
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© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
Question Bank
Section 10.3
Conventional — level 1
2B103C1_001E
Determine whether the following pair of triangles are necessarily similar. If they are, give a proof.
A
D
55
55
65
E
60
B
F
C
Solution
In △ABC,
B + 55 + 60 = 180
( sum of △)
B = 65
In △DEF,
E + 55 + 65 = 180
( sum of △)
E = 60
∵ A = D = 55
B = F = 65
C = E = 60
∴ △ABC ~ △DFE
(AAA)
Answer
no numerical answer
2B103C1_002E
In the figure, AEC is a straight line. Determine whether
A
△ABC and △DEC are necessarily similar. If they are, give a proof.
3 cm
12 cm
E
3 cm
B
Solution
9 cm
4 cm
C 2 cm
D
EC 3 cm 1


BC 9 cm 3
DC
2 cm
1


AC (3  3) cm 3
DE 4 cm 1


AB 12 cm 3
∴
CE DC DE


BC AC AB
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© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
∴ △ABC ~ △DEC
Question Bank
(3 sides proportional)
Answer
no numerical answer
2B103C1_003E
In the figure, ABC is a straight line. Name two similar triangles and give a proof.
D
10.4 cm
E
7.8 cm
8 cm
3 cm
A 2 cm B
C
4 cm
Solution
EA 3 cm 3


BC 4 cm 4
AC (2  4) cm 3


CD
8 cm
4
EC 7.8 cm 3


BD 10.4 cm 4
∴
EA AC EC


BC CD BD
∴ △ACE ~ △CDB
(3 sides proportional)
Answer
no numerical answer
2B103C1_004E
In the figure, ACE and BCD are straight lines. Name two similar triangles and give a proof.
A
B
4 cm
5 cm
C
6 cm
D
7.5 cm
E
Solution
ACB = DCE
(vert. opp. s)
AC 4 cm 2


DC 6 cm 3
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© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
Question Bank
BC
5 cm 2


EC 7.5 cm 3
∴
AC BC

DC EC
∴ △ABC ~ △DEC (ratio of 2 sides, inc. )
Answer
no numerical answer
2B103C1_005E
In the figure, ACD, AGE, FED and BCGF are straight lines.
A
B
AB // FD.
(a) Prove that △ABC ~ △DFC.
C
(b) Prove that △ABG ~ △EFG.
G
F
Solution
(a) ABC = DFC
(alt. s, AB // FD)
BAC = FDC
(alt. s, AB // FD)
ACB = DCF
(vert. opp. s)
∴ △ABC ~ △DFC
(AAA)
(b) ABG = EFG
E
D
(proved in (a))
BAG = FEG
(alt. s, AB // FD)
AGB = EGF
(vert. opp. s)
∴ △ABG ~ △EFG
(AAA)
Answer
(a) no numerical answer
(b) no numerical answer
2B103C1_006E
In the figure, BCDE is a straight line. AB // EF and AD // CF. ADB = 83 and CEF = 45.
Prove that △ABD ~ △FEC.
A
B
83
C
D
45
F
E
Solution
ABD = FEC = 45
(alt. s, AB // EF)
FCE = ADB = 83
(alt. s, AD // CF)
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© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
Question Bank
In △ABD,
BAD + ABD + ADB = 180
( sum of △)
BAD + 45 + 83 = 180
BAD = 52
In △FEC,
EFC + FEC + FCE = 180
( sum of △)
EFC + 45 + 83 = 180
EFC = 52
∴ BAD = EFC
∴ △ABD ~ △FEC
(AAA)
Answer
no numerical answer
2B103C1_007E
In the figure, ABC and CDE are straight lines. AE // BD.
AC = 14 cm, BC = 8 cm and BD = 4 cm.
14 cm
B
A
(a) Prove that △ACE ~ △BCD.
8 cm
C
4 cm
(b) Using the result of (a), find AE.
D
E
Solution
(a) CAE = CBD
(corr. s, BD // AE)
AEC = BDC
(corr. s, BD // AE)
ACE = BCD
(common )
∴ △ACE ~ △BCD
(AAA)
(b) △ACE ~ △BCD
∴
AE
AC
=
BD
BC
AE =
=
(proved in (a))
(corr. sides, ~△s)
14 cm
× BD
8 cm
14
× 4 cm
8
= 7 cm
Answer
(a) no numerical answer
(b) 7 cm
47
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Mathematics in Focus Book 2B Chapter 10
Question Bank
2B103C1_008E
In the figure, ABD and ACE are straight lines. BC // DE.
A
AB = 6 cm, BC = 5 cm and DE = 10 cm.
6 cm
(a) Prove that △ABC ~ △ADE.
(b) Using the result of (a), find AD and BD.
B
5 cm
D
10 cm
C
E
Solution
(a) ABC = ADE
(corr. s, BC // DE)
ACB = AED
(corr. s, BC // DE)
BAC = DAE
(common )
∴ △ABC ~ △ADE
(AAA)
(b) △ABC ~ △ADE
∴
(proved in (a))
DE
AD
=
AB
BC
AD =
(corr. sides, ~△s)
10 cm
× AB
5 cm
= 2 × 6 cm
= 12 cm
BD = AD – AB
= (12 – 6) cm
= 6 cm
Answer
(a) no numerical answer
(b) AD = 12 cm, BD = 6 cm
2B103C1_009E
In the figure, ABC and AED are straight lines. AB = 3 cm, BC = 3 cm, AE = 2 cm, ED = 7 cm and
EB = 2.5 cm.
(a) Prove that △AEB ~ △ACD.
(b) Using the result of (a), find CD.
A
2 cm
E
3 cm
B
3 cm
2.5 cm
7 cm
C
D
48
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Mathematics in Focus Book 2B Chapter 10
Question Bank
Solution
(a) BAE = DAC
(common )
AC (3  3) cm

3
AE
2 cm
AD (2  7) cm

3
AB
3 cm
∴
AC AD

AE AB
∴ △AEB ~ △ACD
(b) △AEB ~ △ACD
∴
CD
AC
=
EB
AE
(ratio of 2 sides, inc. )
(proved in (a))
(corr. sides, ~△s)
CD = 3 × EB
= 3 × 2.5 cm
= 7.5 cm
Answer
(a) no numerical answer
(b) 7.5 cm
2B103C1_010E
In the figure, BCD is a straight line. AB = 9 cm, BC = 6.75 cm,
A
CD = 5.25 cm, DA = 6 cm and AC = x cm.
9 cm
(a) Prove that △ABC ~ △DBA.
6 cm
x cm
(b) Using the result of (a), find the value of x.
B
6.75 cm
C
5.25 cm
D
Solution
(a) ABC = DBA
(common )
CB 6.75 cm 3


AB
9 cm
4
AB
9 cm
9 3

 
DB (6.75  5.25) cm 12 4
∴
CB AB

AB DB
∴ △ABC ~ △DBA
(b) △ABC ~ △DBA
∴
AC
AB
=
DA
DB
(ratio of 2 sides, inc. )
(proved in (a))
(corr. sides, ~△s)
x
3
=
6
4
x = 4.5
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Mathematics in Focus Book 2B Chapter 10
Question Bank
Answer
(a) no numerical answer
(b) 4.5
2B103C1_011E
In the figure, AD = 9 cm, AC = 12 cm and BC = 16 cm. BCA = CAD = 90, DCA = 2x – 2
and ABC = x + 15.
(a) Prove that △ACB ~ △DAC.
(b) Using the result of (a), find the value of x.
9 cm
A
D
12 cm
2x – 2
x + 15
B
16 cm
C
Solution
(a) ACB = DAC = 90
(given)
AC 12 cm 4


DA 9 cm 3
CB 16 cm 4


AC 12 cm 3
∴
AC CB

DA AC
∴ △ACB ~ △DAC
(b) △ACB ~ △DAC
ABC = DCA
∴
(ratio of 2 sides, inc. )
(proved in (a))
(corr. s, ~△s)
x + 15 = 2x – 2
15 + 2 = 2x – x
x = 17
Answer
(a) no numerical answer
(b) 17
2B103C1_012E
In the figure, ACE and BCD are straight lines. AC = DC = 20 cm,
A
B
35
BC = 25 cm and EC = 16 cm. BAC = 35 and DCE = 95.
20 cm
(a) Prove that △ABC ~ △EDC.
(b) Using the result of (a), find CDE.
25 cm
C
20 cm
D
50
95
16 cm
E
© Educational Publishing House Ltd
Mathematics in Focus Book 2B Chapter 10
Question Bank
Solution
(a) ACB = ECD = 95
(vert. opp. s)
AC 20 cm 5


EC 16 cm 4
BC 25 cm 5


DC 20 cm 4
∴
AC BC

EC DC
∴ △ABC ~ △EDC
(b) △ABC ~ △EDC
(ratio of 2 sides, inc. )
(proved in (a))
∴ CAB = CED = 35
(corr. s, ~△s)
In △EDC,
CDE + CED + DCE = 180
( sum of △)
CDE + 35 + 95 = 180
CDE = 50
Answer
(a) no numerical answer
(b) 50
51
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Mathematics in Focus Book 2B Chapter 10
Question Bank
Conventional — level 2
2B103C2_001E
In the figure, ADC is a straight line. BA // DE and AE // BC.
A
E
Prove that △ABC ~ △DEA.
D
B
C
Solution
BAC = EDA
(alt. s, BA // DE)
BCA = EAD
(alt. s, AE // BC)
In △ABC,
ABC + BAC + BCA = 180
( sum of △)
ABC = 180 – BAC – BCA
In △DEA,
DEA + EDA + EAD = 180
( sum of △)
DEA = 180 – EDA – EAD
= 180 – BAC – BCA
∴ ABC = DEA
∴ △ABC ~ △DEA
(AAA)
Answer
no numerical answer
2B103C2_002E
In the figure, AB = 1.5 cm, AD = 4 cm,
A
1.5 cm B
DC = 6 cm, BC = 8 cm and BD = 3 cm.
8 cm
4 cm
(a) Prove that △ABD ~ △BDC.
3 cm
(b) Prove that AB // DC.
D
6 cm
C
Solution
(a)
AB 1.5 cm 1


BD 3 cm 2
AD 4 cm 1


BC 8 cm 2
BD 3 cm 1


DC 6 cm 2
∴
AB AD BD


BD BC DC
∴ △ABD ~ △BDC
(3 sides proportional)
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Mathematics in Focus Book 2B Chapter 10
(b) △ABD ~ △BDC
Question Bank
(proved in (a))
ABD = BDC
(corr. s, ~△s)
∴ AB // DC
(alt. s equal)
Answer
(a) no numerical answer
(b) no numerical answer
2B103C2_003E
In the figure, TSR and UVR are straight lines. TS = SR and UV = VR.
T
U
(a) Prove that △TRU ~ △SRV.
(b) Prove that TU // SV.
S
V
Solution
(a) TRU = SRV
(common )
R
TR 2
 2
SR 1
UR 2
 2
VR 1
∴
TR UR

SR VR
∴ △TRU ~ △SRV
(b) △TRU ~ △SRV
(ratio of 2 sides, inc. ))
(proved in (a))
UTR = VSR
(corr. s, ~△s)
∴ TU // SV
(corr. s equal)
Answer
(a) no numerical answer
(b) no numerical answer
2B103C2_004E
In the figure, ABC and AED are straight lines. AB = 4,
A
2
BC = 2, AE = 2, BE = 3 and CD = x. AEB = ACD.
E
4
(a) Prove that △ABE ~ △ADC.
3
(b) Find x.
B
2
C
Solution
(a) AEB = ACD
x
D
(given)
BAE = DAC
(common )
In △ABE,
ABE + AEB + BAE = 180
( sum of △)
ABE = 180 – AEB – BAE
53
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Mathematics in Focus Book 2B Chapter 10
Question Bank
In △ADC,
ADC + ACD + DAC = 180
( sum of △)
ADC = 180 – ACD – DAC
= 180 – AEB – BAE
∴ ABE = ADC
∴ △ABE ~ △ADC
(AAA)
(b) △ABE ~ △ADC
∴
(proved in (a))
CD
AC
=
EB
AE
(corr. sides, ~△s)
x
42
=
3
2
x =3×3
=9
Answer
(a) no numerical answer
(b) 9
2B103C2_005E
In the figure, BDF and CDE are straight lines. BA // EC
A
and AC // BF.
(a) Prove that △BCD ~ △FED.
C
B
(b) Using the result of (a), prove that △CBA ~ △FED.
3 cm
6 cm
D
5 cm
E
10 cm
F
Solution
(a) CDB = EDF
(vert. opp. s)
BD 6 cm 3


FD 10 cm 5
DC 3 cm 3


DE 5 cm 5
∴
BD DC

FD DE
(ratio of 2 sides, inc. )
∴ △BCD ~ △FED
(b) Consider △CBA and △BCD.
ABC = DCB
(alt. s, BA // DC)
ACB = DBC
(alt. s, AC // BF)
In △CBA,
BAC + ABC + ACB = 180
( sum of △)
BAC = 180 – ABC – ACB
54
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Mathematics in Focus Book 2B Chapter 10
Question Bank
In △BCD,
CDB + DCB + DBC = 180
( sum of △)
CDB = 180 – DCB – DBC
= 180 – ABC – ACB
∴ BAC = CDB
∴ △CBA ~ △BCD
(AAA)
∵ △BCD ~ △FED
(proved in (a))
∴ △CBA ~ △FED
Answer
(a) no numerical answer
(b) no numerical answer
2B103C2_006E
In the figure, ABC and EDC are straight lines. EA  AC
E
x cm
and BD  EC. AB = 3 cm, BC = 6 cm, CD = 4 cm and DE = x cm.
D
(a) Prove that △CDB ~ △CAE.
4 cm
(b) Find the value of x.
A 3 cm B
Solution
(a) BDC = EAC = 90
6 cm
C
(given)
DCB = ACE
(common )
In △CDB,
CBD + BDC + DCB = 180
( sum of △)
CBD + 90 + DCB = 180
CBD = 90 – DCB
In △CAE,
CEA + EAC + ACE = 180
( sum of △)
CEA + 90 + DCB = 180
CEA = 90 – DCB
∴ CBD = CEA
∴ △CDB ~ △CAE
(AAA)
(b) △CDB ~ △CAE
∴
(proved in (a))
CE
CA
=
CB
CD
(corr. sides, ~△s)
( x  4) cm
(6  3) cm
=
6 cm
4 cm
x + 4 = 13.5
x = 9.5
Answer
(a) no numerical answer
(b) 9.5
55
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Mathematics in Focus Book 2B Chapter 10
Question Bank
2B103C2_007E
In the figure, AEC is a straight line. AB = 2.8 cm, BC = 2.4 cm,
A
(a) Prove that △ABC ~ △CDE.
1.4 cm
2.8 cm
(b) Prove that AB // DC and ED // BC.
1 cm
C
Solution
(a)
D
1.2 cm
E
CD = 1.4 cm, DE = 1.2 cm and CE = EA = 1 cm.
2.4 cm
AB 2.8 cm

2
CD 1.4 cm
B
BC 2.4 cm

2
DE 1.2 cm
AC (1  1) cm

2
CE
1 cm
∴
AB BC AC


CD DE CE
∴ △ABC ~ △CDE
(3 sides proportional)
(b) △ABC ~ △CDE
(proved in (a))
BAC = DCE and BCA = DEC.
(corr. s, ~△s)
∴ AB // DC and ED // BC.
(alt. s equal)
Answer
(a) no numerical answer
(b) no numerical answer
2B103C2_008E
In the figure, ABC, AGD, AFE, BGF and CDE are straight lines.
C
AB = 6 cm, BC = 9 cm, AG = 4 cm and GD = 6 cm.
9 cm
(a) Prove that △ABG ~ △ACD.
B
(b) Using the result of (a), prove that BF // CE.
(c) If CEA = 70, findBFE.
6 cm
A
4 cm
G
6 cm
D
F
Solution
(a) BAG = CAD
(common )
E
AB
6 cm
2


AC (6  9) cm 5
AG
4 cm
2


AD (4  6) cm 5
∴
AB AG

AC AD
∴ △ABG ~ △ACD
(ratio of 2 sides, inc. )
56
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Mathematics in Focus Book 2B Chapter 10
Question Bank
(b) △ABG ~ △ACD
(proved in (a))
∴ ABG = ACD
(corr. s, ~△s)
∴ BF // CE
(corr. s equal)
(c) ∵ BF // CE
(proved in (b))
∴ BFE + CEA = 180
(int. s, BF // CE)
BFE + 70 = 180
BFE = 110
Answer
(a) no numerical answer
(b) no numerical answer
(c) 110
2B103C2_009E
In the figure, ABC = CAD and BA // CD.
A
(a) Prove that △ABC ~ △CAD.
(b) Show that AC2 = AB × CD.
D
(c) If AB = 8 cm and CD = 2 cm, find AC.
B
C
Solution
(a) ABC = CAD
(given)
BAC = ACD
(alt. s, BA // CD)
In △ABC,
ACB + BAC + ABC = 180
( sum of △)
ACB = 180 – BAC – ABC
In △CAD,
CDA + ACD + CAD = 180
( sum of △)
CDA = 180 – ACD – CAD
= 180 – BAC – ABC
∴ ACB = CDA
∴ △ABC ~ △CAD
(AAA)
(b) △ABC ~ △CAD
(proved in (a))
AC AB
=
CD CA
∴
(corr. sides, ~△s)
AC2 = AB × CD
(c) AC2 = AB × CD
(proved in (b))
AC = 8 2 cm
= 16 cm
= 4 cm
Answer
(a) no numerical answer
(b) no numerical answer
(c) 4 cm
57
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Mathematics in Focus Book 2B Chapter 10
Question Bank
2B103C2_010E
In the figure, AGDC, FGB and EDB are straight lines.
A
9 cm
2.5 cm F
BAC = FEB and ABC = EFB. AB = 6 cm,
E
G
BC = 5 cm, EF = 9 cm and FG = 2.5 cm.
(a) Prove that △ABC ~ △EFB.
6 cm
D
(b) Using the result of (a), find GB.
(c) Hence, prove that CGB = GCB.
C
B
5 cm
Solution
(a) BAC = FEB
(given)
ABC = EFB
(given)
In △ABC,
ACB + BAC + ABC = 180
( sum of △)
ACB = 180 – BAC – ABC
In △EFB,
EBF + FEB + EFB = 180
( sum of △)
EBF = 180 – FEB – EFB
= 180 – BAC – ABC
∴ ACB = EBF
∴ △ABC ~ △EFB
(AAA)
(b) △ABC ~ △EFB
∴
(proved in (a))
FB
EF
=
BC
AB
FB =
(corr. sides, ~△s)
9
× 5 cm
6
= 7.5 cm
GB = FB – FG
= (7.5 – 2.5) cm
= 5 cm
(c) ∵ GB = CB = 5 cm
∴ CGB = GCB
(base s, isos. △)
Answer
(a) no numerical answer
(b) 5 cm
(c) no numerical answer
58
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Mathematics in Focus Book 2B Chapter 10
Question Bank
Conventional — level 3
2B103C3_001E
In the figure, BCDE is a straight line. DA and CF intersect at G.
BA // CF and DA // EF.
A
(a) Prove that △ABD ~ △GCD.
(b) Prove that
F
GC GD
.

FC FE
G
(c) Using the results of (a) and (b), prove that
B
AB × FE = AD × FC.
C
D
E
Solution
Suggested marks: 8 marks
(a) ABD = GCD
(corr. s, AB // FC)
BAD = CGD
(corr. s, AB // FC)
ADB = GDC
(common )
∴ △ABD ~ △GCD
(AAA)
(b) FCE = GCD
(1M)
(1A)
(common )
FEC = GDC
(corr. s, DA // EF)
CFE = CGD
(corr. s, DA // EF)
∴ △FCE ~ △GCD
(AAA)
(1M)
(corr. sides, ~△s)
(1A)
∴
GC GD

FC FE
(c) △ABD ~ △GCD
∴
GC GD
=
AD
AB
∴
GC
AB
=
AD
GD
∵
GC GD
=
FE
FC
∴
GC
FC
=
GD
FE
∴
AB
FC
=
AD
FE
(1M)
(proved in (a))
(corr. sides, ~△s)
(1M)
(proved in (b))
(1M)
∴ AB × FE = AD × FC
(1A)
Answer
(a) no numerical answer
(b) no numerical answer
(c) no numerical answer
59
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Mathematics in Focus Book 2B Chapter 10
Question Bank
2B103C3_002E
In the figure, △ADE and △BAC are two isosceles triangles with DA = DE and AB = AC.
ABC = x and BAD = CAE = y.
(a) If 3x + 2y = 180, express DAE in terms of x.
(b) Using the result of (a), prove that △ADE ~ △BAC.
(c) Using the result of (b), prove that AD × (BC + AC) = BA × (AE + DE).
A
y
y
D
E
x
C
B
Solution
Suggested marks: 11 marks
(a) In △BAC,
ACB = ABC = x
(base s, isos. △)
ABC + BAC + ACB = 180
( sum of △)
x + (y + DAE + y) + x = 180
(1M)
(1M)
2x + 2y + DAE = 180
∵ 3x + 2y = 180
(given)
∴ 2x +2y + DAE = 3x + 2y
(1M)
DAE = x
(1A)
(b) DEA = DAE = x
(base s, isos. △)
∴ DEA = ACB and DAE = ABC.
In △ADE,
ADE + DEA + DAE = 180
( sum of △)
ADE + x + x = 180
ADE = 180 – 2x
(1M)
In △BAC,
BAC + ACB + ABC = 180
( sum of △)
BAC + x + x = 180
BAC = 180 – 2x
(1M)
∴ ADE = BAC
∴ △ADE ~ △BAC
(AAA)
(c) △ADE ~ △BAC
∴
(1A)
(proved in (b))
AD AE
=
BA BC
(corr. sides, ~△s)
AD × BC = BA × AE .......... (1)
(1M)
60
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Mathematics in Focus Book 2B Chapter 10
Question Bank
AD DE
=
BA AC
(corr. sides, ~△s)
AD × AC = BA × DE .......... (2)
(1M)
(1) + (2): AD × BC + AD × AC = BA × AE + BA × DE
(1M)
AD × (BC + AC) = BA × (AE + DE)
(1A)
Answer
(a) DAE = x
(b) no numerical answer
(c) no numerical answer
61
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Mathematics in Focus Book 2B Chapter 10
Question Bank
Multiple Choice Questions — level 1
2B103M1_001E
Which of the following reasons cannot be used to prove that two triangles are similar?
A. AAA
B.
2 sides proportional
C.
3 sides proportional
D. ratio of 2 sides, inc. 
Solution
2 sides proportional cannot be used to prove that two triangles are similar.
∴ The answer is B.
Answer
B
2B103M1_002E
In the figure, ABC and AED are straight lines. Which of the
A
following reasons can be used to prove △ABE ~ △ACD?
A. AAA
1.1 cm
1.2 cm
B. SAS
B
E
C. 3 sides proportional
1.1 cm
D. ratio of 2 sides, inc. 
C
1.2 cm
D
Solution
BAE = CAD
(common )
AB
1.1 cm
1


AC (1.1  1.1) cm 2
AE
1.2 cm
1


AD (1.2  1.2) cm 2
∴
AB AE

AC AD
∴ △ABE ~ △ACD
(ratio of 2 sides, inc. )
∴ The answer is D.
Answer
D
2B103M1_003E
If △ABC ~ △PQR, which of the following must be true?
I.
AB = PQ
II.
BCA = QRP
III.
AC BC

PR QR
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Mathematics in Focus Book 2B Chapter 10
Question Bank
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
Solution
For I:
The lengths of the corresponding sides of two similar triangles are not necessarily the same.
∴ I may not be true.
For II:
△ABC ~ △PQR
(given)
∴ BCA = QRP
(corr. s, ~△s)
∴ II must be true.
For III:
AC BC

PR QR
(corr. sides, ~△s)
∴ III must be true.
∴ The answer is C.
Answer
C
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Mathematics in Focus Book 2B Chapter 10
Question Bank
Multiple Choice Questions — level 2
2B103M2_001E
In the figure, AMCE and BNCD are straight lines.
A
How many pair(s) of similar triangles is/are there?
B
M
N
A. 1
B.
2
C.
3
C
D. 4
D
E
Solution
△ABC ~ △MNC (AAA)
△ABC ~ △EDC (AAA)
△MNC ~ △EDC (AAA)
∴ There are 3 pairs of similar triangles in the figure.
∴ The answer is C.
Answer
C
2B103M2_002E (TSA)
In the figure, PQ = PS = 6 and QS = RS = 4. Prove that △PQS ~ △SQR.
Q
Which of the following proofs is correct?
A. PQ = PS
(given)
SQ = SR
(given)
QPS = QSR
(given)
∴ △PQS  △SQR
(SAS)
B.
6
R
4
40
4
40
P
6
S
PQ 6 3
 
SQ 4 2
PS 6 3
 
SR 4 2
C.
∴
QS 3

QR 2
∴
PQ PS QS


SQ SR QR
(corr. sides, ~△s)
∴ △PQS ~ △SQR
(3 sides proportional)
QPS = QSR
(given)
PQ 6 3
 
SQ 4 2
PS 6 3
 
SR 4 2
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Mathematics in Focus Book 2B Chapter 10
∴
Question Bank
PQ PS

SQ SR
∴ △PQS  △SQR
D. QPS = QSR
(SAS)
(given)
PQ 6 3
 
SQ 4 2
PS 6 3
 
SR 4 2
∴
PQ PS

SQ SR
∴ △PQS ~ △SQR
(ratio of 2 sides, inc. )
Solution
Since we want to prove that △PQS ~ △SQR, the proofs in choices A and C are not correct.
∴ Choices A and C are not correct.
Since we have not proved that △PQS ~ △SQR, we cannot use the theorem ‘corr. sides, ~ △s’.
∴ The proof in choice B is incorrect.
Only the proof in choice D is correct.
∴ The answer is D.
Answer
D
2B103M2_003E
If △DFE ~ △XYZ and △LMN ~ △YZX, which of the following must be true?
I.
DEF = NML
II.
DE FE

LN MN
III. DF × ML = EF × NL
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
Solution
∵ △DFE ~ △XYZ and △LMN ~ △YZX.
(given)
∴ △DFE ~ △NLM
For I:
△DFE ~ △NLM
∴ DEF = NML
(corr. s, ~△s)
i.e. I must be true.
For II:
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Mathematics in Focus Book 2B Chapter 10
Question Bank
△DFE ~ △NLM
∴
DE
FE

NM LM
∴
DE FE
may not be true.

LN MN
(corr. sides, ~△s)
i.e. II may not be true.
For III:
△DFE ~ △NLM
DF EF
=
NL ML
∴
(corr. sides, ~△s)
DF × ML = EF × NL
i.e. III must be true.
∴ The answer is B.
Answer
B
2B103M2_004E
In the figure, ABC, GHB, EDB, AHFE and CDFG are
G
E
straight lines. BAH = BCD and ABH = CBD.
F
Which of the following must be true?
I.
△ABH ~ △CBD
II.
△AEB ~ △CGB
D
H
III. △GDB ~ △EHB
A
2 cm
B
2 cm
C
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
Solution
For I:
BAH = BCD
(given)
ABH = CBD
(given)
In △ABH,
AHB + BAH + ABH = 180
( sum of △)
AHB = 180 – BAH – ABH
In △CBD,
CDB + BCD + CBD = 180
( sum of △)
CDB = 180 – BCD – CBD
= 180 – BAH – ABH
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Mathematics in Focus Book 2B Chapter 10
Question Bank
∴ AHB = CDB
∴ △ABH ~ △CBD
(AAA)
i.e. I must be true.
For II:
EAB = GCB
(given)
ABE = ABH + GBE
CBG = CBD + GBE
= ABH + GBE
∴ ABE = CBG
In △AEB,
AEB + EAB + ABE = 180
( sum of △)
AEB = 180 – EAB – ABE
In △CGB,
CGB + GCB + CBG = 180
( sum of △)
CGB = 180 – GCB – CBG
= 180 – EAB – ABE
∴ AEB = CGB
∴ △AEB ~ △CGB
(AAA)
i.e. II must be true.
For III:
△AEB ~ △CGB
(proved)
∴ DGB = HEB
(corr. s, ~△s)
GBD = EBH
(common )
In △GDB,
GDB + DGB + GBD = 180
( sum of △)
GDB = 180 – DGB – GBD
In △EHB,
EHB + HEB + EBH = 180
( sum of △)
EHB = 180 – HEB – EBH
= 180 – DGB – GBD
∴ GDB = EHB
∴ △GDB ~ △EHB
(AAA)
i.e. III must be true.
∴ The answer is D.
Answer
D
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Mathematics in Focus Book 2B Chapter 10
Question Bank
Section 10.4
Conventional — level 1
2B104C1_001E
Copy the following angle by using a pair of compasses and a straight edge only.
Solution
Answer
no numerical answer
2B104C1_002E
Copy the following angle by using a pair of compasses and a straight edge only.
Solution
Answer
no numerical answer
2B104C1_003E
Construct the perpendicular bisector of the following line segment by using
a pair of compasses and a straight edge only.
D
E
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Mathematics in Focus Book 2B Chapter 10
Question Bank
Solution
D
E
Answer
no numerical answer
2B104C1_004E
Construct the perpendicular bisector of the following line segment by using
a pair of compasses and a straight edge only.
A
B
Solution
A
B
Answer
no numerical answer
2B104C1_005E
Construct the angle bisector of the following angle by using a pair of compasses
and a straight edge only.
Solution
Answer
no numerical answer
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Mathematics in Focus Book 2B Chapter 10
Question Bank
2B104C1_006E
Construct the angle bisector of the following angle by using a pair of compasses
and a straight edge only.
Solution
Answer
no numerical answer
2B104C1_007E
Construct the angle bisector of the following angle by using a pair of compasses
and a straight edge only.
Solution
Answer
no numerical answer
2B104C1_008E
Construct an angle of size 90.
Solution
Answer
no numerical answer
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Mathematics in Focus Book 2B Chapter 10
Question Bank
2B104C1_009E
Construct an angle of size 60.
Solution
60
Answer
no numerical answer
2B104C1_010E
Construct an angle of size 22.5.
Solution
22.5
Answer
no numerical answer
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Mathematics in Focus Book 2B Chapter 10
Question Bank
Conventional — level 2
2B104C2_001E
Construct an angle of size 225 by using a pair of compasses and a straight edge only.
Solution
225 = 180 + 45
225
Answer
no numerical answer
2B104C2_002E
Construct an angle of size 67.5 by using a pair of compasses and a straight edge only.
Solution
67.5 = 45 + 22.5
67.5
Answer
no numerical answer
2B104C2_003E
Construct an angle of size 330 by using a pair of compasses and a straight edge only.
Solution
330 = 360 – 30
330
Answer
no numerical answer
2B104C2_004E
Divide the line segment AB into 4 equal parts by using a pair of compasses and a straight
edge only.
A
B
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Mathematics in Focus Book 2B Chapter 10
Question Bank
Solution
A
B
Answer
no numerical answer
2B104C2_005E
Divide AOB into 4 equal angles by using a pair of compasses and a straight edge only.
A
B
O
Solution
A
B
O
Answer
no numerical answer
2B104C2_006E
Construct a regular hexagon by using a pair of compasses and a straight edge only.
Solution
Answer
no numerical answer
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Mathematics in Focus Book 2B Chapter 10
Question Bank
2B104C2_007E
Construct a regular pentagon by using a pair of compasses and a straight edge only.
Solution
Answer
no numerical answer
2B104C2_008E
Construct a triangle which is congruent to △PQR by using a pair of compasses
and a straight edge only.
Q
R
P
Solution
Answer
no numerical answer
2B104C2_009E
Construct a triangle which is congruent to △XYZ by using a pair of compasses
and a straight edge only.
X
Y
Z
Solution
Answer
no numerical answer
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Mathematics in Focus Book 2B Chapter 10
Question Bank
2B104C2_010E
Construct an isosceles triangle with base angles 45 by using a pair of compasses
and a straight edge only.
Solution
45
45
Answer
no numerical answer
2B104C2_011E
(a) Construct the angle bisector of each interior angle of △ABC by using a
pair of compasses and a straight edge only.
B
C
A
(b) What do you find about the angle bisectors constructed in (a)?
Solution
(a)
B
C
A
(b) The angle bisectors constructed in (a) meet at one point.
Answer
(a) no numerical answer
(b) they meet at one point
2B104C2_012E
(a) Construct the perpendicular bisector of each side of △PQR by using a
pair of compasses and a straight edge only.
Q
P
R
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Mathematics in Focus Book 2B Chapter 10
Question Bank
(b) What do you find about the perpendicular bisectors constructed in (a)?
Solution
(a)
Q
P
R
(b) The perpendicular bisectors constructed in (a) meet at one point.
(a) no numerical answer
(b) they meet at one point
2B104C2_013E
In the figure, A is a point lying on straight line L and B is a point lying above L.
B
L
A
By using a pair of compasses and a straight edge only,
(a) construct a straight line L1 passing through A and perpendicular to L.
(b) Using the result of (a), construct a straight line L2 passing through B and parallel to L.
Solution
(a)
B
L
A
L1
(b)
B
C
L2
L
A
L1
(a) no numerical answer
(b) no numerical answer
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Mathematics in Focus Book 2B Chapter 10
Question Bank
Multiple Choice Questions — level 1
2B104M1_001E
When we construct the angle bisector of AOB, after drawing an arc to cut AO and OB at
D and E respectively, we
A. take D and E as the centres, draw two arcs with the same radius so that they intersect at F.
B.
take D and E as the centres, draw two arcs with the different radius so that they intersect at F.
C.
take A and B as the centres, draw two arcs with the same radius so that they intersect at F.
D. take A and B as the centres, draw two arcs with the different radius so that they intersect at F.
Solution
In order to draw OF which is the angle bisector of AOB, we take D and E as the centres, draw two
arcs with the same radius so that they intersect at F.
∴ The answer is A.
Answer
A
2B104M1_002E
When constructing the perpendicular bisector of a line segment AB, we first take A as the
centre and a suitable radius to draw arcs above and below AB.
The suitable radius cannot be
I.
shorter than half of AB.
II.
longer than half of AB.
III. equal to half of AB.
IV. as long as AB.
A. I and II only
B.
C.
D. I, II and III only
II and III only
I and III only
Solution
In order to have a point of intersection between two arcs, the suitable radius must be longer than
half of AB.
∴ The suitable radius can neither be shorter than half of AB nor be equal to half of AB.
∴ The answer is B.
Answer
B
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Mathematics in Focus Book 2B Chapter 10
Question Bank
Multiple Choice Questions — level 2
2B104M2_001E
Which of the angles of the following sizes can be constructed by using a pair of compasses
and a straight edge only?
I.
165
II.
255
III. 292.5
A. II only
B.
C.
D. I, II and III
I and III only
III only
Solution
For I:
165 = 180 – 15
For II:
255 = 180 + 45 + 30
For III:
292.5 = 180 + 90 + 22.5
∴ I, II and III must be true.
∴ The answer is D.
Answer
D
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