IEB practice · Paper II

IEB Euclidean geometry practice

19 newly generated questions across all 14 euclidean geometry topics the IEB examines in Paper II, 219 marks in all. Work each one, then open its answer underneath to check yourself. Nothing here is taken from a past paper.

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The tangent-chord theorem

2 questions · 24 marks

Paper II · Section A · 13 marks

QUESTION 1

In the diagram:

A, B, C and T lie on the circle.
PTQ is a tangent to the circle at T.
AB ∥ PQ.
TC cuts AB at D.
Q $\hat{T}$ B = 66°

State all relevant reasons with your statements.

(a)
Complete the table: give the reason for each statement, in the order given.
Statement Reason
T $\hat{A}$ B = 66°
T $\hat{B}$ A = 66°
TA = TB
A $\hat{T}$ B = 48°
A $\hat{C}$ B = 132°
T $\hat{C}$ B = 66°
A $\hat{C}$ T = 66°
(7)
(b)
Hence prove that TC bisects A $\hat{C}$ B.
(1)
(c)
Prove that ΔTDB ||| ΔTBC.
(3)
(d)
Hence prove that TB^{^2} = TD · TC.
(2)

Statement	Reason
T $\hat{A}$ B = 66°
T $\hat{B}$ A = 66°
TA = TB
A $\hat{T}$ B = 48°
A $\hat{C}$ B = 132°
T $\hat{C}$ B = 66°
A $\hat{C}$ T = 66°

[13]

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(a) Answer: Each reason is given beside its statement below.

T $\hat{A}$ B = 66° (tan chord theorem)
T $\hat{B}$ A = 66° (alt ∠s; PQ ∥ AB)
TA = TB (sides opp equal ∠s)
A $\hat{T}$ B = 48° (sum ∠s of Δ)
A $\hat{C}$ B = 132° (opp ∠s of cyclic quad)
T $\hat{C}$ B = 66° (∠s in the same seg)
A $\hat{C}$ T = 66° (∠s in the same seg)

(b) Answer: A $\hat{C}$ T = T $\hat{C}$ B

A $\hat{C}$ T = T $\hat{C}$ B = 66° (proved)

(c) Answer: ΔTDB ||| ΔTBC

In ΔTDB and ΔTBC:
D $\hat{T}$ B = B $\hat{T}$ C (common)
T $\hat{B}$ D = T $\hat{C}$ B = 66° (proved)
∴ ΔTDB ||| ΔTBC (∠∠∠)

(d) Answer: TB^{^2} = TD · TC

TB/TD = TC/TB (||| Δs)
∴ TB^{^2} = TD · TC

Paper II · Section B · 11 marks

QUESTION 2

In the diagram:

Two circles touch each other externally at T.
MTN is the common tangent to the two circles at T.
ATB and CTD are straight lines, with A and C on the one circle and B and D on the other.
C $\hat{A}$ T = x

State all relevant reasons with your statements.

(a)
State, with reasons, three other angles equal to x.
(4)
(b)
Hence prove that AC ∥ DB.
(2)
(c)
Prove that ΔTAC ||| ΔTBD.
(3)
(d)
If TA = 10 units, TB = 20 units and TC = 8 units, calculate the length of TD.
(2)

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(a) Answer: ${\hat{T}}_{2}$ , ${\hat{T}}_{5}$ and T $\hat{B}$ D

${\hat{T}}_{2}$ = C $\hat{A}$ T = x (tan chord theorem)
${\hat{T}}_{5}$ = ${\hat{T}}_{2}$ = x (vert opp ∠s =)
T $\hat{B}$ D = ${\hat{T}}_{5}$ = x (tan chord theorem)

(b) Answer: AC ∥ DB

C $\hat{A}$ T = T $\hat{B}$ D = x (proved)
∴ AC ∥ DB (alt ∠s =)

(c) Answer: ΔTAC ||| ΔTBD

In ΔTAC and ΔTBD:
${\hat{T}}_{3}$ = ${\hat{T}}_{6}$ (vert opp ∠s =)
C $\hat{A}$ T = T $\hat{B}$ D = x (proved)
∴ ΔTAC ||| ΔTBD (∠∠∠)

(d) Answer: TD = 16 units

TA/TB = TC/TD (||| Δs)
∴ TD = (8 · 20)/10 = 16 units

The angle at the centre

1 questions · 9 marks

Paper II · Section A · 9 marks

QUESTION 3

In the diagram:

O is the centre of the circle.
A, B, D and T lie on the circle.
PTQ is a tangent to the circle at T.
T $\hat{B}$ A = 46°

State all relevant reasons with your statements.

(a)
Determine P $\hat{T}$ A.
(2)
(b)
Determine T $\hat{O}$ A.
(2)
(c)
Determine O $\hat{T}$ A.
(2)
(d)
Determine T $\hat{D}$ A.
(2)
(e)
Determine the size of reflex T $\hat{O}$ A.
(1)

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(a) Answer: P $\hat{T}$ A = 46°

P $\hat{T}$ A = T $\hat{B}$ A = 46° (tan chord theorem)

(b) Answer: T $\hat{O}$ A = 92°

T $\hat{O}$ A = 2 × T $\hat{B}$ A = 92° (∠ at centre = 2 × ∠ at circumference)

(c) Answer: O $\hat{T}$ A = 44°

O $\hat{T}$ P = 90° (tan ⊥ radius)
∴ O $\hat{T}$ A = 90° − P $\hat{T}$ A = 44°

(d) Answer: T $\hat{D}$ A = 134°

T $\hat{D}$ A = 180° − T $\hat{B}$ A = 134° (opp ∠s of cyclic quad)

(e) Answer: reflex T $\hat{O}$ A = 268°

reflex T $\hat{O}$ A = 360° − 92° = 268° (∠s round a pt)

Tangents, and tangents from a point outside

1 questions · 14 marks

Paper II · Section B · 14 marks

QUESTION 4

In the diagram:

O is the centre of the circle.
PA and PB are tangents to the circle at A and B respectively.
C is a point on the circle such that AC ∥ PB.
A $\hat{C}$ B = 39°

State all relevant reasons with your statements.

(a)
Determine P $\hat{B}$ A.
(2)
(b)
Determine A $\hat{P}$ B.
(3)
(c)
Is ΔABC an isosceles triangle? Explain.
(4)
(d)
Prove that PAOB is a cyclic quadrilateral.
(3)
(e)
Determine the size of A $\hat{O}$ B.
(2)

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(a) Answer: P $\hat{B}$ A = 39°

P $\hat{B}$ A = A $\hat{C}$ B = 39° (tan chord theorem)

(b) Answer: A $\hat{P}$ B = 102°

PA = PB (tans from common pt)
P $\hat{A}$ B = P $\hat{B}$ A = 39° (∠s opp equal sides)
∴ A $\hat{P}$ B = 102° (sum ∠s of Δ)

(c) Answer: Yes: BA = BC, so ΔABC is isosceles

C $\hat{A}$ B = A $\hat{B}$ P = 39° (alt ∠s; AC ∥ PB)
A $\hat{C}$ B = 39° (given)
∴ C $\hat{A}$ B = A $\hat{C}$ B
∴ BA = BC (sides opp equal ∠s)

(d) Answer: PAOB is a cyclic quadrilateral

O $\hat{A}$ P = 90° (tan ⊥ radius)
O $\hat{B}$ P = 90° (tan ⊥ radius)
∴ O $\hat{A}$ P + O $\hat{B}$ P = 180°
∴ PAOB is a cyclic quad (opp ∠s of quad supp)

(e) Answer: A $\hat{O}$ B = 78°

A $\hat{O}$ B = 2 × A $\hat{C}$ B = 78° (∠ at centre = 2 × ∠ at circumference)

Proving the theorem itself, not using it

6 questions · 74 marks

Paper II · Section A · 13 marks

QUESTION 5

In the diagram:

O is the centre of the circle, and SAT is a tangent to the circle at A.
B and C lie on the circle.

State all relevant reasons with your statements.

(a)
Use the diagram to prove the theorem that states that the angle between a tangent to a circle and a chord drawn from the point of contact is equal to the angle in the alternate segment.
(6)
In the diagram below, A, B and C lie on the circle and PAQ is a tangent to the circle at A. D lies on AB and E on AC, with DE ∥ BC.
(b)
Prove that P $\hat{A}$ D = A $\hat{E}$ D.
(3)
(c)
Hence prove that PQ is a tangent to the circle through A, D and E.
(2)
(d)
If AD = 3 units, DB = 1 units and AC = 8 units, calculate the length of AE.
(2)

[13]

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(a) Answer: See the proof below.

Required to prove: B $\hat{A}$ T = A $\hat{C}$ B
Construction: Draw the diameter AD and join DB.
D $\hat{A}$ T = 90° (tan ⊥ radius)
A $\hat{B}$ D = 90° (∠ in semi-circle)
∴ B $\hat{A}$ T = 90° − D $\hat{A}$ B
A $\hat{D}$ B = 90° − D $\hat{A}$ B (sum ∠s of Δ)
∴ B $\hat{A}$ T = A $\hat{D}$ B
A $\hat{D}$ B = A $\hat{C}$ B (∠s in the same seg)
∴ B $\hat{A}$ T = A $\hat{C}$ B

(b) Answer: P $\hat{A}$ D = A $\hat{E}$ D

P $\hat{A}$ B = A $\hat{C}$ B (tan chord theorem)
A $\hat{C}$ B = A $\hat{E}$ D (corresp ∠s; DE ∥ BC)
∴ P $\hat{A}$ D = A $\hat{E}$ D

(c) Answer: PQ is a tangent to the circle through A, D and E

P $\hat{A}$ D = A $\hat{E}$ D (proved)
∴ PQ is a tangent to the circle through A, D and E (converse tan chord theorem)

(d) Answer: AE = 6 units

AE/EC = AD/DB = 3 (line ∥ one side of Δ; DE ∥ BC)
∴ AE = (3/4) × 8 = 6 units

Paper II · Section A · 14 marks

QUESTION 6

In the diagram:

In ΔABC and ΔDEF, $\hat{A}$ = $\hat{D}$ , $\hat{B}$ = $\hat{E}$ and $\hat{C}$ = $\hat{F}$ .

State all relevant reasons with your statements.

(a)
Use the diagram to prove the theorem that states that if two triangles are equiangular, their corresponding sides are in proportion.
(6)
In the diagram below, ΔPQR has a right angle at Q, and S lies on PR with QS ⊥ PR. PS = 9 units and SR = 16 units.
(b)
Prove that ΔPQS ||| ΔQRS.
(3)
(c)
Hence prove that QS^{^2} = PS · SR.
(2)
(d)
Calculate the lengths of QS and PQ.
(3)

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(a) Answer: See the proof below.

Required to prove: DE/AB = DF/AC
Construction: Mark G on AB so that AG = DE, and H on AC so that AH = DF; join GH.
In ΔAGH and ΔDEF:
AG = DE (construction)
G $\hat{A}$ H = E $\hat{D}$ F (given)
AH = DF (construction)
∴ ΔAGH ≡ ΔDEF (SAS)
∴ A $\hat{G}$ H = D $\hat{E}$ F = A $\hat{B}$ C (≡ Δs)
∴ GH ∥ BC (corresp ∠s =)
AG/AB = AH/AC (line ∥ one side of Δ; GH ∥ BC)
∴ DE/AB = DF/AC

(b) Answer: ΔPQS ||| ΔQRS

In ΔPQS and ΔQRS:
P $\hat{S}$ Q = Q $\hat{S}$ R = 90° (given)
Q $\hat{P}$ S = 90° − P $\hat{Q}$ S (sum ∠s of Δ)
P $\hat{Q}$ R = 90° (given)
∴ R $\hat{Q}$ S = 90° − P $\hat{Q}$ S
∴ ΔPQS ||| ΔQRS (∠∠∠)

(c) Answer: QS^{^2} = PS · SR

PS/QS = QS/SR (||| Δs)
∴ QS^{^2} = PS · SR

(d) Answer: QS = 12 units; PQ = 15 units

QS^{^2} = 9 · 16 = 144, so QS = 12 units (proved)
∴ PQ^{^2} = 9^{^2} + 12^{^2} = 225, so PQ = 15 units (Pythagoras)

Paper II · Section A · 13 marks

QUESTION 7

In the diagram:

O is the centre of the circle, and A, B and P lie on the circle.

State all relevant reasons with your statements.

(a)
Use the diagram to prove the theorem that states that the angle subtended by an arc at the centre of a circle is twice the angle the same arc subtends at the circle.
(6)
In the diagram below, O is the centre of the circle and ABCD is a cyclic quadrilateral. B $\hat{O}$ D = 4x and B $\hat{C}$ D = 2x + 60°.
(b)
Express B $\hat{A}$ D in terms of x.
(2)
(c)
Hence determine the value of x.
(3)
(d)
Determine the size of O $\hat{B}$ D.
(2)

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(a) Answer: See the proof below.

Required to prove: A $\hat{O}$ B = 2 × A $\hat{P}$ B
Construction: Join PO and produce it to Q.
OA = OP (radii)
${\hat{P}}_{1}$ = O $\hat{A}$ P (∠s opp equal sides)
${\hat{O}}_{1}$ = 2 × ${\hat{P}}_{1}$ (ext ∠ of Δ)
${\hat{O}}_{2}$ = 2 × ${\hat{P}}_{2}$ (ext ∠ of Δ; similarly)
∴ A $\hat{O}$ B = 2 × ${\hat{P}}_{1}$ + 2 × ${\hat{P}}_{2}$
∴ A $\hat{O}$ B = 2 × A $\hat{P}$ B

(b) Answer: B $\hat{A}$ D = 2x

B $\hat{A}$ D = ½ × B $\hat{O}$ D = 2x (∠ at centre = 2 × ∠ at circumference)

(c) Answer: x = 30°

2x + 2x + 60° = 180° (opp ∠s of cyclic quad)
4x = 120°
∴ x = 30°

(d) Answer: O $\hat{B}$ D = 30°

OB = OD (radii)
O $\hat{B}$ D = O $\hat{D}$ B (∠s opp equal sides)
∴ O $\hat{B}$ D = 30° (sum ∠s of Δ)

Paper II · Section A · 12 marks

QUESTION 8

In the diagram:

ABCD is a cyclic quadrilateral in the circle with centre O.

State all relevant reasons with your statements.

(a)
Use the diagram to prove the theorem that states that the opposite angles of a cyclic quadrilateral are supplementary.
(5)
In the diagram below, ABCD is a cyclic quadrilateral. AB is produced to E and DC is produced to F so that EF ∥ BC.
(b)
Prove that A $\hat{E}$ F = A $\hat{B}$ C.
(1)
(c)
Hence prove that ADFE is a cyclic quadrilateral.
(4)
(d)
If A $\hat{D}$ C = 73°, determine the size of A $\hat{E}$ F.
(2)

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(a) Answer: See the proof below.

Required to prove: B $\hat{A}$ D + B $\hat{C}$ D = 180°
Construction: Join OB and OD.
${\hat{O}}_{1}$ = 2 × B $\hat{A}$ D (∠ at centre = 2 × ∠ at circumference)
reflex B $\hat{O}$ D = 2 × B $\hat{C}$ D (∠ at centre = 2 × ∠ at circumference)
${\hat{O}}_{1}$ + reflex B $\hat{O}$ D = 360° (∠s round a pt)
∴ 2 × B $\hat{A}$ D + 2 × B $\hat{C}$ D = 360°
∴ B $\hat{A}$ D + B $\hat{C}$ D = 180°

(b) Answer: A $\hat{E}$ F = A $\hat{B}$ C

A $\hat{E}$ F = A $\hat{B}$ C (corresp ∠s; BC ∥ EF)

(c) Answer: ADFE is a cyclic quadrilateral

A $\hat{B}$ C + A $\hat{D}$ C = 180° (opp ∠s of cyclic quad)
A $\hat{E}$ F = A $\hat{B}$ C (proved)
∴ A $\hat{E}$ F + A $\hat{D}$ F = 180°
∴ ADFE is a cyclic quad (opp ∠s of quad supp)

(d) Answer: A $\hat{E}$ F = 107°

A $\hat{B}$ C = 180° − A $\hat{D}$ C = 107° (opp ∠s of cyclic quad)
∴ A $\hat{E}$ F = 107° (proved)

Paper II · Section A · 10 marks

QUESTION 9

In the diagram:

O is the centre of the circle and AB is a chord.
OM ⊥ AB, with M on AB.

State all relevant reasons with your statements.

(a)
Use the diagram to prove the theorem that states that the line drawn from the centre of a circle perpendicular to a chord bisects the chord.
(5)
In the diagram below, O is the centre of the circle, and KL and NQ are parallel chords on opposite sides of O. KL = 8 units, NQ = 6 units, and EOF ⊥ KL with E on KL and F on NQ. The chords are 7 units apart.
(b)
Calculate the radius of the circle. (Let OE = y.)
(5)

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(a) Answer: See the proof below.

Required to prove: AM = MB
Construction: Join OA and OB.
In ΔOMA and ΔOMB:
OA = OB (radii)
OM = OM (common)
O $\hat{M}$ A = O $\hat{M}$ B = 90° (given)
∴ ΔOMA ≡ ΔOMB (RHS)
∴ AM = MB (≡ Δs)

(b) Answer: r = 5 units

KE = EL = 4 and NF = FQ = 3 (line from centre ⊥ to chord)
r^{^2} = 4^{^2} + y^{^2} (Pythagoras)
r^{^2} = 3^{^2} + (7 − y)^{^2} (Pythagoras)
16 + y^{^2} = 9 + 49 − 14y + y^{^2}
∴ y = 3
∴ r = √(16 + 9) = 5 units

Paper II · Section A · 12 marks

QUESTION 10

In the diagram:

In ΔABC, D lies on AB and E on AC with DE ∥ BC.

State all relevant reasons with your statements.

(a)
Use the diagram to prove the theorem that states that a line drawn parallel to one side of a triangle divides the other two sides proportionally.
(6)
In the diagram below, S lies on PQ and T on PR with ST ∥ QR, and U lies on QR with SU ∥ PR. PS = 3 units, SQ = 2 units and QR = 10 units.
(b)
Write down PT : TR.
(1)
(c)
Determine QU : UR.
(2)
(d)
Calculate the length of ST.
(3)

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(a) Answer: See the proof below.

Required to prove: AD/DB = AE/EC
Construction: Join BE and CD.
area ΔADE/area ΔBDE = AD/DB (same height)
area ΔADE/area ΔCED = AE/EC (same height)
area ΔBDE = area ΔCED (same base; same height; DE ∥ BC)
∴ AD/DB = AE/EC

(b) Answer: PT : TR = 3 : 2

PT : TR = PS : SQ = 3 : 2 (line ∥ one side of Δ; ST ∥ QR)

(c) Answer: QU : UR = 2 : 3

In ΔQPR: SU ∥ PR
∴ QU : UR = 2 : 3 (line ∥ one side of Δ; SU ∥ PR)

(d) Answer: ST = 6 units

UR = (3/5) × 10 = 6 units (QU : UR = 2 : 3)
STRU is a parallelogram (both pairs opp sides ∥)
∴ ST = UR = 6 units (opp sides of pgm)

Cyclic quadrilaterals

1 questions · 12 marks

Paper II · Section B · 12 marks

QUESTION 11

In the diagram:

Two circles cut each other at A and B.
P lies on the first circle; PA produced meets the second circle at Q, and PB produced meets it at R.
XPY is a tangent to the first circle at P.
X $\hat{P}$ A = x

State all relevant reasons with your statements.

(a)
Express P $\hat{B}$ A in terms of x.
(2)
(b)
Express A $\hat{Q}$ R in terms of x.
(2)
(c)
Prove that XY ∥ QR.
(2)
(d)
Prove that ΔPAB ||| ΔPRQ.
(3)
(e)
If PA = 4 units, AQ = 5 units and PB = 3 units, calculate the length of BR.
(3)

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(a) Answer: P $\hat{B}$ A = x

P $\hat{B}$ A = X $\hat{P}$ A = x (tan chord theorem)

(b) Answer: A $\hat{Q}$ R = x

A $\hat{Q}$ R = P $\hat{B}$ A = x (ext ∠ of cyclic quad)

(c) Answer: XY ∥ QR

X $\hat{P}$ Q = P $\hat{Q}$ R = x (proved)
∴ XY ∥ QR (alt ∠s =)

(d) Answer: ΔPAB ||| ΔPRQ

In ΔPAB and ΔPRQ:
A $\hat{P}$ B = R $\hat{P}$ Q (common)
P $\hat{B}$ A = P $\hat{Q}$ R = x (proved)
∴ ΔPAB ||| ΔPRQ (∠∠∠)

(e) Answer: BR = 9 units

PA/PR = PB/PQ (||| Δs)
∴ PR = (4 · 9)/3 = 12 units
∴ BR = 12 − 3 = 9 units

Similar triangles

2 questions · 21 marks

Paper II · Section A · 10 marks

QUESTION 12

In the diagram:

ABCD is a trapezium with AB ∥ DC.
The diagonals AC and BD intersect at E.
F lies on AD and G on BC, with FEG ∥ AB.
AB = 18 units and DC = 9 units.

State all relevant reasons with your statements.

(a)
Prove that ΔABE ||| ΔCDE.
(3)
(b)
Hence write down AE : EC.
(1)
(c)
Calculate the length of FE.
(3)
(d)
Prove that E is the midpoint of FG.
(3)

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(a) Answer: ΔABE ||| ΔCDE

In ΔABE and ΔCDE:
B $\hat{A}$ E = D $\hat{C}$ E (alt ∠s; AB ∥ DC)
A $\hat{B}$ E = C $\hat{D}$ E (alt ∠s; AB ∥ DC)
∴ ΔABE ||| ΔCDE (∠∠∠)

(b) Answer: AE : EC = 2 : 1

AE : EC = 2 : 1 (||| Δs)

(c) Answer: FE = 6 units

In ΔADC: FE ∥ DC
ΔAFE ||| ΔADC (∠∠∠)
FE/DC = AE/AC = 2/3 (||| Δs)
∴ FE = (2/3) × 9 = 6 units

(d) Answer: FE = EG, so E is the midpoint of FG

In ΔBDC: EG ∥ DC
EG/DC = BE/BD = 2/3 (||| Δs)
∴ EG = (2/3) × 9 = 6 units
∴ FE = EG

Paper II · Section B · 11 marks

QUESTION 13

In the diagram:

B, C, D and E lie on the circle.
A is a point outside the circle; ADB and AEC are straight lines.
DE and BC are joined.

State all relevant reasons with your statements.

(a)
Prove that ΔADE ||| ΔACB.
(4)
(b)
Hence prove that AD · AB = AE · AC.
(2)
(c)
If AD = 4 units, DB = 8 units and AE = 3 units, calculate the length of EC.
(3)
(d)
If, further, DE = 3 units, calculate the length of BC.
(2)

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(a) Answer: ΔADE ||| ΔACB

In ΔADE and ΔACB:
D $\hat{A}$ E = C $\hat{A}$ B (common)
${\hat{D}}_{1}$ = A $\hat{C}$ B (ext ∠ of cyclic quad)
∴ A $\hat{E}$ D = A $\hat{B}$ C (sum ∠s of Δ)
∴ ΔADE ||| ΔACB (∠∠∠)

(b) Answer: AD · AB = AE · AC

AD/AC = AE/AB (||| Δs)
∴ AD · AB = AE · AC

(c) Answer: EC = 13 units

4 · 12 = 3 · AC (proved)
∴ AC = 16 units
∴ EC = 16 − 3 = 13 units

(d) Answer: BC = 12 units

DE/CB = AD/AC (||| Δs)
∴ BC = (3 · 16)/4 = 12 units

Products and ratios of lengths

1 questions · 8 marks

Paper II · Section A · 8 marks

QUESTION 14

In the diagram:

In ΔABC, D lies on BC with BD : DC = 4 : 5.
E is the midpoint of AD, and BE produced meets AC at F.
G lies on FC such that DG ∥ BF.

State all relevant reasons with your statements.

(a)
Determine FG : GC.
(2)
(b)
Prove that AF = FG.
(2)
(c)
Hence determine AF : FC.
(2)
(d)
Determine area ΔABF : area ΔCBF.
(2)

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(a) Answer: FG : GC = 4 : 5

In ΔCBF: DG ∥ BF
∴ FG : GC = 4 : 5 (line ∥ one side of Δ; DG ∥ BF)

(b) Answer: AF = FG

In ΔADG: AE = ED and EF ∥ DG
∴ AF = FG (line through midpt ∥ to second side)

(c) Answer: AF : FC = 4 : 9

AF = FG = 4k and GC = 5k (proved)
∴ AF : FC = 4 : 9

(d) Answer: 4 : 9

area ΔABF : area ΔCBF = 4 : 9 (same height)

The proportion theorem

1 questions · 10 marks

Paper II · Section A · 10 marks

QUESTION 15

In the diagram:

In ΔABC, D lies on AB and E lies on AC with DE ∥ BC.
AD : DB = 3 : 1.
BE and CD are joined.

State all relevant reasons with your statements.

(a)
Write down AE : EC, giving a reason.
(2)
(b)
Determine area ΔADE : area ΔBDE.
(2)
(c)
Prove that area ΔBDE = area ΔCDE.
(2)
(d)
Determine the numerical value of (area ΔADE)/(area ΔABC).
(4)

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(a) Answer: AE : EC = 3 : 1

AE : EC = 3 : 1 (line ∥ one side of Δ; DE ∥ BC)

(b) Answer: 3 : 1

area ΔADE : area ΔBDE = 3 : 1 (same height)

(c) Answer: area ΔBDE = area ΔCDE

area ΔBDE = area ΔCDE (same base; same height; DE ∥ BC)

(d) Answer: 9/16

area ΔADE/area ΔADC = 3/4 (same height)
area ΔADC/area ΔABC = 3/4 (same height)
∴ area ΔADE/area ΔABC = 9/16

The angle in a semi-circle

1 questions · 13 marks

Paper II · Section B · 13 marks

QUESTION 16

In the diagram:

AB is a diameter of the circle with centre O.
The tangent to the circle at C meets AB produced at D.
${\hat{C}}_{3}$ = x

State all relevant reasons with your statements.

(a)
Express C $\hat{A}$ B in terms of x.
(2)
(b)
Express ${\hat{C}}_{2}$ in terms of x.
(3)
(c)
Prove that ΔDCB ||| ΔDAC.
(3)
(d)
Hence prove that DC^{^2} = DB · DA.
(2)
(e)
If DB = 7 units and DC = 14 units, calculate the radius of the circle.
(3)

[13]

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(a) Answer: C $\hat{A}$ B = x

C $\hat{A}$ B = ${\hat{C}}_{3}$ = x (tan chord theorem)

(b) Answer: ${\hat{C}}_{2}$ = 90° − x

A $\hat{C}$ B = 90° (∠ in semi-circle)
${\hat{C}}_{1}$ = C $\hat{A}$ O = x (∠s opp equal sides)
∴ ${\hat{C}}_{2}$ = 90° − x

(c) Answer: ΔDCB ||| ΔDAC

In ΔDCB and ΔDAC:
C $\hat{D}$ B = A $\hat{D}$ C (common)
${\hat{C}}_{3}$ = C $\hat{A}$ B = x (proved)
∴ ΔDCB ||| ΔDAC (∠∠∠)

(d) Answer: DC^{^2} = DB · DA

DC/DA = DB/DC (||| Δs)
∴ DC^{^2} = DB · DA

(e) Answer: r = 10.5 units

14^{^2} = 7(7 + 2r) (proved)
196 = 49 + 14r
∴ r = 10.5 units

Chords, and the line from the centre

1 questions · 10 marks

Paper II · Section A · 10 marks

QUESTION 17

In the diagram:

A, B, C and D lie on the circle.
AB = BC.
EBF is a tangent to the circle at B.
B $\hat{A}$ C = 25°

State all relevant reasons with your statements.

(a)
Prove that DB bisects A $\hat{D}$ C.
(2)
(b)
Prove that EF ∥ AC.
(4)
(c)
Determine the size of A $\hat{B}$ C.
(2)
(d)
Determine the size of ${\hat{D}}_{1}$ .
(2)

[10]

Show the answers

(a) Answer: ${\hat{D}}_{1}$ = ${\hat{D}}_{2}$ , so DB bisects A $\hat{D}$ C

AB = BC (given)
∴ ${\hat{D}}_{1}$ = ${\hat{D}}_{2}$ (equal chords; equal ∠s)

(b) Answer: EF ∥ AC

E $\hat{B}$ A = B $\hat{C}$ A (tan chord theorem)
B $\hat{C}$ A = B $\hat{A}$ C (∠s opp equal sides)
∴ E $\hat{B}$ A = B $\hat{A}$ C
∴ EF ∥ AC (alt ∠s =)

(c) Answer: A $\hat{B}$ C = 130°

B $\hat{C}$ A = B $\hat{A}$ C = 25° (∠s opp equal sides)
∴ A $\hat{B}$ C = 130° (sum ∠s of Δ)

(d) Answer: ${\hat{D}}_{1}$ = 25°

${\hat{D}}_{1}$ = A $\hat{C}$ B = 25° (∠s in the same seg)

Proving a quadrilateral is cyclic

1 questions · 15 marks

Paper II · Section B · 15 marks

QUESTION 18

In the diagram:

In ΔABC, D lies on BC and E lies on AC.
AD ⊥ BC and BE ⊥ AC.
AD and BE intersect at H.
$\hat{C}$ = x

State all relevant reasons with your statements.

(a)
Prove that CDHE is a cyclic quadrilateral.
(3)
(b)
Prove that ABDE is a cyclic quadrilateral.
(3)
(c)
Express D $\hat{A}$ C in terms of x.
(2)
(d)
Prove that ${\hat{D}}_{1}$ = B $\hat{A}$ C.
(2)
(e)
Hence prove that ΔCDE ||| ΔCAB.
(3)
(f)
If CD = 12 units, CA = 20 units and CE = 9 units, calculate the length of CB.
(2)

[15]

Show the answers

(a) Answer: CDHE is a cyclic quadrilateral

H $\hat{D}$ C = 90° (given)
H $\hat{E}$ C = 90° (given)
∴ H $\hat{D}$ C + H $\hat{E}$ C = 180°
∴ CDHE is a cyclic quad (opp ∠s of quad supp)

(b) Answer: ABDE is a cyclic quadrilateral

A $\hat{D}$ B = A $\hat{E}$ B = 90° (given)
∴ ABDE is a cyclic quad (line subtends equal ∠s)

(c) Answer: D $\hat{A}$ C = 90° − x

A $\hat{D}$ C = 90° (given)
∴ D $\hat{A}$ C = 90° − x (sum ∠s of Δ)

(d) Answer: ${\hat{D}}_{1}$ = B $\hat{A}$ C

${\hat{D}}_{1}$ = B $\hat{A}$ E (ext ∠ of cyclic quad)

(e) Answer: ΔCDE ||| ΔCAB

In ΔCDE and ΔCAB:
D $\hat{C}$ E = A $\hat{C}$ B (common)
${\hat{D}}_{1}$ = C $\hat{A}$ B (proved)
∴ ΔCDE ||| ΔCAB (∠∠∠)

(f) Answer: CB = 15 units

CD/CA = CE/CB (||| Δs)
∴ CB = (9 · 20)/12 = 15 units

The midpoint theorem

1 questions · 9 marks

Paper II · Section A · 9 marks

QUESTION 19

In the diagram:

In quadrilateral ABCD, P, Q, R and S are the midpoints of AB, BC, CD and DA respectively.
The diagonals AC and BD are drawn.

State all relevant reasons with your statements.

(a)
Prove that PQ ∥ AC and PQ = ½AC.
(2)
(b)
Hence prove that PQRS is a parallelogram.
(3)
(c)
If AC = 14 units and BD = 19 units, calculate the perimeter of PQRS.
(2)
(d)
Determine area ΔAPS : area ΔABD.
(2)

[9]

Show the answers

(a) Answer: PQ ∥ AC and PQ = ½AC

In ΔABC: AP = PB and BQ = QC
∴ PQ ∥ AC (midpt theorem)
∴ PQ = ½AC (midpt theorem)

(b) Answer: PQRS is a parallelogram

In ΔADC: DS = SA and DR = RC
SR ∥ AC (midpt theorem)
SR = ½AC (midpt theorem)
∴ PQ ∥ SR and PQ = SR
∴ PQRS is a parallelogram (one pair opp sides = and ∥)

(c) Answer: 33 units

PQ = SR = ½(14) = 7 (midpt theorem)
PS = QR = ½(19) = 9.5 (midpt theorem)
∴ perimeter = 14 + 19 = 33 units

(d) Answer: 1 : 4

area ΔAPS : area ΔABS = 1 : 2 (same height)
area ΔABS : area ΔABD = 1 : 2 (same height)
∴ area ΔAPS : area ΔABD = 1 : 4

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