主要供中三學生參考 四邊形 (Quadrilaterals) - V3 - Final - G
主要供中三學生參考 四邊形 (Quadrilaterals) - V3 - Final - G
主要供中三學生參考 四邊形 (Quadrilaterals) - V3 - Final - G
1. Introduction to quadrilaterals
A polygon is a closed plane figure which is enclosed by three or more line segments.
A quadrilateral is a polygon with four sides. The following are some common
quadrilaterals.
Trapezium
Isosceles
Trapezium
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Some common terms relating to a quadrilateral
Interior angle
Side
Diagonal
Vertex
(a) Vertex (頂點): the points where the adjacent sides meet, e.g. A, B, C and D.
(b) Side:
(i) Sides (邊): the line segments of the polygon, e.g. AB, BC, CD and AD.
(ii) Adjacent sides (鄰邊): two sides with a common vertex,
eg. AB and BC are a pair of adjacent sides. B is the common vertex of the sides
(iii) Opposite sides (對邊): two sides with no common vertex,
eg. AB and DC are a pair of opposite sides.
(c) Angle:
(i) Angle (角) : when two lines meet at a point, they form an angle, e.g. ∠ABC.
(ii) Adjacent angles (鄰角): two angles with a common side,
eg. ∠ABC and ∠BCD are a pair of adjacent angles . BC is the common side of
the angles.
(iii) Opposite angles (對角): two angles with no common side,
eg. ∠BAD and∠BCD are a pair of opposite angles.
(d) Diagonal (對角線) : the line segment joining any two non-adjacent vertices of a
polygon, e.g. AC .
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2. Parallelograms
2.1 Definition
A parallelogram (平行四邊形) is a quadrilateral with two pairs of parallel opposite sides.
In the figure, since BC//AD and BA//CD, the quadrilateral ABCD is a parallelogram.
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2.3 Conditions for Parallelograms
Conditions Figures
1 Both pairs of opposite sides of a quadrilateral
are equal.
i.e. If BA=CD and BC = AD ,
then ABCD is a parallelogram.
[Reference: opp. sides equal ]
2 Both pairs of opposite angles of a quadrilateral
are equal.
i.e. If ∠A =∠C and ∠B =∠D,
then ABCD is a parallelogram.
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Example: In the figure, ABCD is a parallelogram. Find the values of m and n.
n
3.5
Solution
∵ ABCD is a parallelogram.
∴ AB = DC (opp. sides of // gram)
m=6
4m
Solution
∵ ABCD is a parallelogram.
∴ ∠C = ∠A (opp. ∠s of // gram)
4m = 80°
m = 20°
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Example: In the figure, ABCD is a parallelogram. The diagonals AC and BD intersect
at O. Find the values of m and n.
n
2m
8
6
Solution
∵ ABCD is a parallelogram.
∴ OD = OB (diags. of // gram)
2m = 8
m=4
2m
4n
Solution
∵ ABCD is a parallelogram.
∴ ∠C =∠A (opp. ∠s of // gram)
2m = 80°
m = 40°
∠A +∠B = 180° (int. ∠s, DA // CB)
80° + 4n = 180°
n = 25°
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Example: The figure shows a parallelogram ABCD with perimeter 38 cm.
AB =12 cm. Find the length of BC.
12
Solution
∵ ABCD is a parallelogram.
∵ DC = AB (opp. sides of // gram)
= 12 cm
∵ Perimeter of ABCD = 38 cm
∴ AB + BC + CD + AD = 38
12 + BC + 12 + BC = 38
2BC = 14
BC = 7 cm
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Example: The figure shows a quadrilateral ABCD .
Prove that ABCD is a parallelogram.
12
10
10
12
Solution
∵ AB = DC = 12
and AD = BC = 10
∴ ABCD is a parallelogram. opp. sides equal
110° 70°
70° 110°
Solution
∵ A = C = 70
and B = D = 110
∴ ABCD is a parallelogram. opp. s equal
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Example: In the figure, DB and AC intersect at O.
Prove that ABCD is a parallelogram.
6
8
8
6
Solution
∵ AO = OC = 6
and BO = OD = 8
∴ ABCD is a parallelogram. diags. bisect each other
5 cm 5 cm
70° 110°
Solution
∵ A + C = 70 +110
= 180
∴ DA // CB int. s supp.
∵ DA = CB = 5 cm
and DA // CB
∴ ABCD is a parallelogram. opp. sides equal and //
Page 9 of 57
3. Rhombuses (菱形)
3.1 Introduction
A rhombus is a quadrilateral with four equal sides.
OR
A rhombus is a parallelogram with one pair of equal adjacent sides.
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3.2 Properties of rhombuses
(1) All properties of a parallelogram.
(2) All sides are equal.
DB ⊥ CA
3.3 Note
(a) When you join the midpoint of the sides of a rhombus, you will get a rectangle.
(b) The two diagonals of a rhombus NOT equal.
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Example: In the figure, ABCD is a rhombus. Find the value of m.
(3m+6) cm 36
cm
cm
Solution
∵ ABCD is a rhombus.
∴ BA = BC (definition of rhombus)
3m + 6 = 36
m =10
O
cm
Solution
∵ ABCD is a rhombus.
∴ DO = OB (property of rhombus)
6 = 3n
n=2
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Example:
(a) In the figure, ABCD is a rhombus. ∠ABC = 60. AC = 6 cm.
Find the length of AB and BC.
Solution
(a) ∵ ABCD is a rhombus.
∴ BA = BC (definition of rhombus)
∠BAC =∠BCA (base ∠s, isos. △)
In △ABC,
∠BAC +∠BCA +∠ABC = 180° (∠ sum of △)
∠BAC +∠BAC + 60° = 180°
2∠BAC = 120°
∠BAC = 60°
∴ ∠BAC =∠BCA =∠ABC = 60°
∴ AB=BC=AC
∵ AC = 6 cm
∴ AB = 6 cm and BC =6 cm
BO = 2 √3
BD = 2BO = 2 x 2 √3 = 4√3 cm
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4. Rectangles (長方形, 矩形)
4.1 Introduction
A rectangle is a quadrilateral with four equal interior angles.
OR
A rectangle is a parallelogram with one interior angle equal to right angle (90°).
rectangle.
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4.2 Properties of rectangles
(3) Diagonals are equal. (i.e. both diagonals have the same length).
AC=BD
OA=OB=OC=OD
4.3 Note
(a) A parallelogram with ONE interior right angle is a rectangle.
(b) The diagonals bisect each other at different angles. One is acute, and another
one is an obtuse angle
(c) A rectangle with side lengths m and n has the perimeter as 2(m+n) units
(d) A rectangle with side lengths m and n has the area as mn square units
(e) If m and n are the side lengths of a rectangle, then the length of each diagonal
is √𝑚2 + 𝑛2 units.
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Example: In the figure, ABCD is a rectangle. Find the values of m and n.
m+15°
2n cm 5 cm
cm cm
Solution
∵ ABCD is a rectangle.
∴ AD = BC (property of rectangle)
2n = 5
n = 2.5
and CBA = 90 (definition of rectangle)
m+15° = 90°
m = 75°
Solution
∵ ABCD is a rectangle.
∴ BO = DO (property of rectangle)
q = 10
and AC = BD (property of rectangle)
AC = BO + DO
5p = 10 + 10
p=4
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Example: In the figure, ABCD is a rectangle. The diagonals AC and BD intersect at O.
(a) Find ∠OCB
(b) Find the value of m.
A D
O
m
22
B C
Solution
(a) ∵ ABCD is a rectangle.
∴ OC = OB (property of rectangle)
∴ ∠OCB =∠OBC (base ∠s, isos. △)
= 22°
(b) In △OBC, ∠DOC =∠OBC + ∠OCB (ext. ∠ of △)
m = 22° + 22°
= 44°
A D
E F
O
B C
Solution
∵ ABCD is a rectangle.
∴ OB = OC (property of rectangle)
∠BEO =∠CFO (given)
∠BOE =∠COF (vert. opp. ∠s)
∴ △BOE △COF (AAS)
∴ BE=CF (corr. sides, △s )
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5. Squares (正方形)
5.1 Introduction
A square is a quadrilateral with four equal sides and four equal interior angles.
OR
A square is a parallelogram with one pair of equal adjacent sides and one interior
angle equal to right angle (90°).
A square is also a rhombus and a rectangle. Therefore, squares have all the properties
of rhombuses and rectangles.
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5.2 Properties of squares
(1) All properties of a rhombus.
(2) All properties a rectangle.
(3) Angle between a diagonal and a side is 45°.
5.3 Note
(a) A square with side lengths m has the perimeter as 4m units
(d) A square with side lengths m has the area as 𝑚2 square units
(e) If m is the side length of a square, then the length of each diagonal
is √2𝑚 units.
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Example: In the figure, ABCD is a square. Find the values of p and q.
2p +6
30 cm
q+30°
Solution
∵ ABCD is a square.
∴ D = 90 (definition of square)
q+30° = 90°
q = 60°
3q+15°
cm
Solution
∵ ABCD is a square.
∴ DAE = 45 ( property of square)
p = 45°
Solution
∵ ABCD is a square.
In △CED,
∠CED = 90° (property of square)
CD2 = CE 2 + DE 2 (Pyth. theorem)
CD2 = 102 + 102
CD = 102 + 102
= 14.14 cm (cor. to 4 sig. fig.)
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Example: In the figure, ABCD is a square. AGE, BGF, DFC and BEC are straight lines.
∠AGB = 90°.
(a) Prove that BAE = CBF.
(b) Prove that AE=BF.
A D
F
G
B C
E
Solution
In △ABG,
∠ABG +∠BAG +∠AGB = 180° (∠ sum of △)
∠ABG +∠BAG + 90° = 180°
∠ABG +∠BAG = 90°
i.e. ∠ABG +∠BAE = 90°
∵ ABCD is a square
∴ ABG +CBF = 90 (definition of square)
∴ BAE = CBF
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6. Trapeziums (梯形)
AD : upper base
BC: lower base
h
h = height
In a trapezium.
(a) the two parallel sides are called the bases of the trapezium.
(b) the perpendicular distance between the two bases is the height of the trapezium.
In particular, if the non-parallel opposite sides of a trapezium are equal, then the
trapezium is called an isosceles trapezium (等腰梯形).
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7. References (參考資料): Kites
7.1 A kite (箏形,鳶形) is a quadrilateral that has two distinct pairs of equal adjacent
sides. It has reflectional symmetry about one of the diagonals (but not both).
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8. Properties of the quadrilaterals – an overview
Properties of
quadrilaterals Parallelogram Rhombus Rectangle Square
Diagonals are
perpendicular to each
other No Yes No Yes
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Quadrilaterals Figures
Parallelogram
Rhombus
Rectangle
Square
Page 26 of 57
9. Relationship among different kind of quadrilaterals
The following diagram shows the relationship among different kind of quadrilaterals
we have discussed in this section.
Trapezium
Isosceles
Trapezium
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10. Mid-point Theorem and Intercept Theorem
Page 28 of 57
10.2 Intercept Theorem (截線定理)
M transversal
A 𝐿1
intercept
B
𝐿2
intercept
C
𝐿3
N
In the figure,
(i) the transversal (截線) MN cuts three straight lines 𝐿1 , 𝐿2 and 𝐿3 at points A, B and
C respectively.
(ii) AB is called the intercept (截距) made by 𝐿1 and 𝐿2 on MN , while BC is called the
intercept made by 𝐿2 and 𝐿3 on MN.
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10.2.2 Intercept theorem
Intercept theorem
When three or more parallel lines make
equal intercepts on a given transversal,
these parallel lines will also make equal
intercepts on any other transversal.
i.e. If 𝐿1 ∥𝐿2 ∥𝐿3 and PQ=QR,
then KM=MN.
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Example: In the figure, ADB and AEC are straight lines. DE = 20 cm. Find the
length of BC.
Solution
∵ AD=DB and AE = EC (given)
∴ DE = ½ BC (mid-pt. theorem)
20 = ½ BC
BC = 40 cm
Solution
∵ AD = DB and AE = EC given
∴ DE // FC mid-pt. theorem
∵ AD = DB and BF = FC given
∴ DF // EC mid-pt. theorem
∵ DE // FC and DF // EC proved
∴ CEDF is a parallelogram.
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Example: In the figure, E, F and G are the mid-points of AB, AC and DC respectively.
AEB, AFC, DGC and EFG are straight lines. Prove that AD // BC.
A D
E G
F
B C
Solution
In △ABC,
∵ AE = EB and AF = FC given
∴ EF // BC mid-pt. theorem
i.e. EG // BC (1)
In △ACD,
∴ FG // AD mid-pt. theorem
i.e. EG // AD (2)
∴ AD // BC from (1) and (2)
Page 32 of 57
Example: In the figure, PQR and STU are straight lines. If TU = 10 cm, find the
length of ST.
Solution
∵ PS // QT // RU and PQ = QR
∴ ST = TU (intercept theorem)
= 10 cm
Example: In the figure, ADB and AEC are straight lines. EC = 20 cm.
Find the length of AE.
Solution
∵ AD = DB and DE // BC
∴ AE = EC (intercept theorem)
= 20 cm
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Example: In the figure, KXL and JYM are straight lines. Find the value of p.
(2p+30) cm
5p cm
Solution
∵ KJ // XY // LM and KX = XL
∴ JY = YM (intercept theorem)
2p+30 = 5p
P=10
Page 34 of 57
Part 2: Quiz (1)
3m
3m-3
m+9
4n+4
Solution
∵ ABCD is a parallelogram.
∴ BC = AD (opp. sides of // gram)
3m-3 = m+9
m=6
and AB = DC (opp. sides of // gram)
3m = 4n+4
3(6) = 4n+4
n= 3.5
3m+20°
4.5m-10°
Solution
∵ ABCD is a parallelogram.
∴ ∠C =∠A (opp. ∠s of // gram)
4.5m - 10° = 3m + 20°
m = 20°
Page 35 of 57
3. In the figure, ABCD is a parallelogram. The diagonals AC and BD intersect at O.
Find the values of m and n.
n-3
m+2
6
5
Solution
∵ ABCD is a parallelogram.
∴ OD = OB (diags. of // gram)
m+2 = 6
m=4
and OC = OA (diags. of // gram)
n-3 = 5
n=8
A D
22
B
C
Solution
In △ABC,
∵ AC = BC
∴ ∠BAC =∠B (base s, isos. △)
∠BAC +∠B +∠ACB = 180° (∠ sum of △)
∠B +∠B + 22° = 180°
∠B = 79°
∵ ABCD is a parallelogram.
∴ ∠ADC =∠B (opp. ∠s of // gram)
= 79°
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5. In the figure, ABCD is a parallelogram with perimeter 64 cm.
E is a point on DC such that AE ⊥ DC. AE = 12 cm and BC = 14 cm.
Solution
(a) ∵ ABCD is a parallelogram.
∴ AD = BC (opp. sides of // gram)
= 14 cm
and AB = DC (opp. sides of // gram)
Page 37 of 57
6. In the figure, DFE, ABE and CFB are straight lines. CF = BF.
ABCD is a parallelogram.
(a) Prove that △CDF≌△BEF
(b) Prove that AB=BE.
Solution
(a) ∵ ABCD is a parallelogram. (given)
∴ AB//DC (definition of // gram)
i.e. AE//DC
and AB = CD (opp. sides of // gram)
∴ ∠C=∠FBE and ∠CDF=∠E (alt. ∠s, AE // DC)
Page 38 of 57
Part 2: Quiz (2)
Solution
∵ ABCD is a rhombus
1
∴ AO = AC (property of rhombus)
2
1
= 60
2
= 30 cm
In △AOB,
AOB = 90 (property of rhombus)
AO + BO = AB
2 2 2
(Pyth. theorem)
302 + BO2 = 342
BO2 = 256
BO = 16 cm
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2. In the figure, WXYZ is a rhombus. XEW, XFZ and YFE are straight lines.
XF = EF and ∠WXY = 120.
(a) Find ∠EXF .
(b) Find ∠ZYE.
Solution
(a) ∵ WXYZ is a rhombus.
= ½ ∠WXY
= ½ x 120
= 60
i.e. ∠EXF =60
(b) ∵ XF = EF
∴ ∠XEF =∠EXF (base ∠s, isos. )
= 60
∠ZYE = ∠XEF (alt. ∠s, ZY // WX)
= 60°
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3. In the figure, ABCD is a rectangle. The diagonals AC and BD intersect at O.
AB = 60 cm and BC = 80 cm
(a) Find the length of AC
(b) Find the length of OD.
Solution
(a) ∵ ABCD is a rectangle.
∴ ∠ABC = 90° (definition of rectangle)
In △ABC,
∴ OD = OC
= ½ AC
= ½ x 100
= 50 cm
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4. In the figure, ABCD is a square. The diagonals BD and AC intersect at E.
BE = 10 cm.
(a) Find the length of AE
(b) Find the length of AB.
(Leave your answers in surd form if necessary.)
Solution
(a) ∵ ABCD is a square.
∴ AE= BE (property of square)
= 10
(b) In △ABE,
∠AEB = 90° (property of square)
AB2 = BE 2 + AE 2 (Pyth. theorem)
AB2 = 102 + 102
AB = 10 2
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5. In the figure, ABCD is a square. AMD, CME, CDF and AEF are straight lines.
∠CEF= 90°.
(a) Prove that ∠ECF =∠DAF.
(b) Prove that MD = FD
(c) Prove that ∠DFM=45°.
B A
M E
C F
D
Solution
(a) In △CEF,
∠ECF +∠EFC +∠CEF = 180° (∠ sum of △)
∠ECF +∠EFC + 90° = 180°
∠ECF +∠EFC = 90° (1)
∵ ABCD is a square
∴ ADF = 90 (definition of square)
In △ADF,
∠DAF +∠AFD +∠ADF = 180° (∠ sum of △)
∠DAF +∠AFD + 90° = 180°
∠DAF +∠AFD = 90°
i.e. ∠DAF +∠EFC = 90° (2)
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(b) In △CDM and △ADF,
∵ ∠MCD =∠FAD [proved in (a)]
CDM = ADF = 90 (definition of square)
CD=AD (definition of square)
∴ △CDM≌△ADF (ASA)
∴ MD = FD (corr. sides, △s )
Page 44 of 57
Part 2: Quiz (3)
1. In the figure, ADB and AEC are straight lines. ACB = 90. Find AED.
C
2 cm
E
2 cm
A 3 cm D 3 cm B
Solution
∵ AE=EC= 2 cm and AD=DB = 3 cm (given)
∴ DE // BC (mid-pt. theorem)
∴ AED =ACB (corr. ∠s, DE // BC)
= 90
2. In the figure, ADB and AEC are straight lines. ∠ABC = 45°. ∠BAC = 75°.
Find ∠AED.
Solution
In △ABC,
∵ AD = DB and AE = EC (given)
∴ DE // BC (mid-pt. theorem)
∴ ADE = ABC (corr. s, DE // BC)
= 45
In △ADE,
AED + ADE + DAE = 180 ( sum of △)
Page 45 of 57
3. In the figure, ADEB and AFGC are straight lines. DF = 5 cm
(a) Find the length of EG.
(b) Find the length of BC.
A
4.5 cm 3 cm
D F
4.5 cm 3 cm
E G
9 cm 6 cm
B C
Solution
(a) In △AEG,
∵ AD = DE = 4.5 and AF = FG = 3 (given)
∴ DF = ½ EG (mid-pt. theorem)
5 = ½ EG
EG = 10 cm
(b) In △ABC,
AE = AD + DE
= 4.5 +4.5
= 9 cm
AG = AF + FG
=3+3
= 6 cm
∵ AE = EB = 9 and AG = GC = 6
∴ EG = ½ BC (mid-pt. theorem)
10 = ½ BC
BC = 20 cm
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4. In the figure, D, G and E are the mid-points of AB, AF and AC respectively.
CF = DG and BF = 50 cm.
(a) Find the length of CF.
(b) Find the length of GE.
Solution
(a) In △ABF,
∵ AD = DB and AG = GF (given)
∴ DG = ½ BF (mid-pt. theorem)
= ½ x 50
= 25 cm
CF = DG (given)
= 25 cm
(b) In △AFC,
∵ AE = EC and AG = GF (given)
∴ GE = ½ CF (mid-pt. theorem)
= ½ x 25
= 12.5 cm
Page 47 of 57
5. In the figure, AEC, BED and BFC are straight lines. E is the mid-point of both AC
and BD, while F is the mid-point of BC. Prove that AB = DC.
A D
C
B F
Solution
In △ABC,
∴ EF = ½ AB mid-pt. theorem
AB = 2 EF (1)
In △BCD,
∵ BE = ED and BF = FC given
∴ EF = ½ DC mid-pt. theorem
DC = 2 EF (2)
Page 48 of 57
6. In the figure, ABC and DEF are straight lines. AD//BE//CF.
AB =BC = 9.5 cm and EF = 10 cm. Find the length of DF.
Solution
∵ AD // BE // CF and AB =BC
∴ DE = EF (intercept theorem)
= 10 cm
DF = DE + EF
= 10 + 10
= 20 cm
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7. In the figure, ABCM and DEFN are straight lines. AD//BE//CF//MN.
AB=BC=CM =9.6 cm and FN = 9.7 cm. Find the length of DE.
M N 𝑙4
Solution
∵ BE // CF // MN and BC=CM
∴ EF = FN (intercept theorem)
= 9.7 cm
∵ AD // BE // CF and AB = BC
∴ DE = EF (intercept theorem)
= 9.7 cm
Page 50 of 57
8. In the figure, ADFB and AEGC are straight lines. DE // FG // BC.
AD = DF = FB and AC = 90 cm. Find the length of AE.
Solution
In △AFG,
∵ AD = DF and DE // FG
∴ AE = EG (intercept theorem)
∵ DF = FB and DE // FG // BC
∴ EG = GC (intercept theorem)
∴ AE = EG = GC
AE = 1/3 AC
= 1/3 x 90
= 30 cm
Page 51 of 57
Part 3: Challenging Questions (思考題)
(*=Level 1, **=Level 2, ***=Level 3)
**1. In the figure, E, F and G are the mid-points of AB, DC and AC respectively.
AEB, AGC, DFC and EGF are straight lines. AD = 21 cm and BC = 12 cm.
Find the length of EF.
**2. In the figure, E, F and G are the mid-points of AB, DC and AC respectively.
AEB, AGC, DFC and EGF are straight lines. AD = 21 cm and EF = 16.5 cm.
Find the length of BC.
**3. In the figure, E, F and G are the mid-points of AB, DC and AC respectively.
AEB, AGC, DFC and EGF are straight lines. Prove that AD + BC = 2EF.
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**4. In the figure, ADB and AEC are straight lines. AD = DB and AE = EC.
Find the ratio of area of △DEB to area of △EBC.
**5. In the figure, E, G and F are the mid-points of AD, AC and BC respectively.
AED, AGC, BFC and EGF are straight lines.
If EG : GF = 7 : 6, prove that AB : DC = 6 : 7.
**6. In the figure, AED, AGC, BFC and EGF are straight lines. AE = ED.
AB // EF // DC. DC = 40 cm, FC = 8 cm and EF = 30 cm.
(a) Find the lengths of BC,
(b) Find the lengths of GF.
(c) Find the lengths of AB.
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**7. In the figure, AED, BFC, GAB, DCH and GEOFH are straight lines.
ABCD is a parallelogram. AE=CF.
(a) Prove that △ABE≌△CDF.
(b) Join GD. If DG=BG,
(i) prove that BEDF is a parallelogram
(ii) prove that ∠GOB=∠GOD.
**8. In the figure, DEF, CBF, DCH and AEB are straight lines. AE = BE.
ABCD is a parallelogram.
(a) Prove that △ADE≌△BFE.
(b) Join CE. If ∠ADE=∠CDE , prove that ∠CED=∠CEF.
Page 54 of 57
**9. In the figure, AGD, DEC, BFC and GCP are straight lines.
ABCD is a parallelogram. DG=DC and CE=CF.
(a) Prove that ∠DCP=∠FCP.
(b) Prove that FP=EP.
**10. In the figure, AOCM, EOF, DFC and BEC are straight lines. AE = AF.
ABCD is a square.
(a) Prove that BE = DF.
(b) If OM = OA, prove that AEMF is a rhombus.
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***12. In the figure, AED, BDC and BEF are straight lines. ∠BAC=90°.
BD = CD = AD. AE= DE. AF//BC.
(a) Prove that △AEF≌△DEB.
(b) Prove that ADCF is a rhombus.
(c) If AB=5 cm and AC=4 cm,find the area of ADCF.
***13. In the figure, AFC and BCD are straight lines. ∠ACB=90°.
EA⊥AB. EF⊥AC,ED⊥BC. AB=AE. BC=4 cm. FC=5 cm.
(a) Prove that △ABC≌△EAF. (b) Find the length of AC and EF.
(c) Find the length of AB and AE. (d) Prove that EFCD is a rectangle.
(e) Find the perimeter of quadrilateral ABDE.
***14. In the figure, BFC, AMC,FMD and CED are straight lines.
ABCD is a rhombus. ME⊥CD. BF=CF. ∠1=∠2.
(a) Prove that MC=MD.
(b) Prove that △MCE≌△MDE.
(c) If CE=1 cm,
(i) find the length of BC.
(ii) prove that MF=ME.
(iii) prove that AM=DF+ME.
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***15. Refer to below Figure (M) and Figure (N)
(1) Figure (M): ABCD is a square. AED and DFC are straight lines. ∠EBF=∠CBF.
Produce DC such that CG=AE. Join BG.
(i) Prove that △ABE≌△CBG.
(ii) Prove that ∠ABF=∠FBG .
(iii) Prove that BE = AE + CF.
A E D F
A
B
F
B C D C E
H
G
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