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© Gauteng Department of Education
1
SECONDARY SCHOOL IMPROVEMENT
PROGRAMME (SSIP) 2019
GRADE 12
SUBJECT: MATHEMATICS
LEARNER NOTES (PAGE 1 OF 15)
© Gauteng Department of Education
2
D
E F
G H
NO: 5
TOPIC: EUCLIDEAN GEOMETRY (GRADE 12 TRIANGLE GEOMETRY)
SECTION A: TYPICAL EXAM QUESTIONS
QUESTION 1
1.1 In DEF , GH||EF. DG :GE 5:3 . DE 32 mm and DF 24 mm.
Determine the length of DG, GE, DH and HF. (6)
1.2 In ABC , CD 14 cm, DA 6 cm, CE 21 cm and EB 9 cm .
Prove that DE||BC. (5)
1.3 In ACE , BF||CE, 38
BC AC and AE : ED 4:3 .
Determine DG :GB . (6)
QUESTION 2
PB is a tangent to circle ABC. PA||BC. Prove that:
2.1 PAB||| ABC (5)
2.2 PA:AB AB:BC (2)
2.3 2PA . BC AB (2)
2.4 AP AB
BP AC (2)
A
B
C
D
E
F
G
© Gauteng Department of Education
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A B
CD
E
12
1 2
1
212
O
C
A
B
1 2
1
212
1 2
D1
2 3
4
QUESTION 3
ABCD is a cyclic quadrilateral. AB and DC produced
meet at E.
Prove that: AE AD
CE BC (7)
QUESTION 4
ABCD is a cyclic quadrilateral, AC and BD intersect at P.
E is a point on BD such that AE||DC.
Prove that:
4.1 AP PE
PC PD (6)
4.2 2AP BP . PE (9)
QUESTION 5
In trapezium ABCD, DC 2BC ,
1ˆ ˆA E and BC EC .
Prove that:
5.1 AD BD
EC DC (6)
5.2 BD 2AD (3)
QUESTION 6
ABOC is a kite in which ˆB C 90 .
6.1 Why is OCD||| OAC ? (2)
6.2 Hence complete:
6.2.1 2OC ..... ..... (1)
6.2.2 2CA ..... ..... (1)
6.2.3 2CD ..... ..... (1)
© Gauteng Department of Education
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4 parts 3 parts
21cm
A C B
6.3 Prove that:
6.3.1 2
2
BD AD
AOOB (6)
6.3.2 2 2OC OD OD. DA (2)
6.4 If 12
OD DA x , prove that CD 2 . OD (3)
SECTION B: NOTES ON CONTENT
REVISION OF THE CONCEPT OF RATIOS
Consider the line segment AB. If AB 21 cm and C divides AB in the ratio AC:CB 4:3 , it
is possible to find the actual lengths of AC and CB.
It is clear that AC doesn’t equal 4cm and CB doesn’t equal 3 cm because 4 3 21 cm .
However, if we let each part equal k, it will be possible to find the length of AC and CB in
centimetres.
The length of AC is (4 )cmk and the length of CB is (3 )cmk .
4 3 21 cm
7 21 cm
3 cm
k k
k
k
Each part represents 3 cm.
AC 4(3 cm) 12 cm
and CB 3(3 cm) 9 cm
4k 3k
A C B
k k k k k k k
A C B
3 cm 3 cm 3 cm 3 cm 3 cm 3 cm 3 cm
12 cm 9 cm
© Gauteng Department of Education
5
A
B C
D E
hk
Note:
AC 12cm 4AC:CB
CB 9cm 3
4:3 is the ratio of AC:CB.
THEOREM
A line drawn parallel to one side of a
triangle cuts the other two sides so as
to divide them in the same proportion.
If DE||BC then AD AE
DB EC
Proof
In ADE , draw height h relative to base AD and height k relative to base AE.
Join BE and DC to create BDE and CED . 12
12
. AD .Area ADE AD
Area BDE . BD . BD
h
h
12
12
. AE .Area ADE AE
Area CED . EC . EC
k
k
Now it is clear that
Area BDE Area CED
(same base, height and
lying between parallel lines)
Area ADE Area ADE
Area BDE Area CED
AD AE
BD EC
Corollaries
(1) AB AC
AD AE (2)
AB AC
DB EC (3)
BD CE
DA EA (4)
BD CE
BA CA
Whenever you use this theorem the reason you must give is: Line || one side of
© Gauteng Department of Education
6
A
B C
D E
THEOREM CONVERSE
If a line cuts two sides of a triangle proportionally,
then that line is parallel to the third side.
If AD AE
DB EC then DE||BC
Whenever you use this theorem the
reason you must give is:
Line divides sides of proportionally
THEOREM (MIDPOINT THEOREM)
The line passing through the midpoint of
one side of a triangle, parallel to another
side, bisects the third side and is equal to
half the length of the side it is parallel to.
If AD DB and DE||BC , then AE EC
and 12
BC 2DE or DE BC .
Also, if AD DB and AE EC , then
DE||BC and 12
BC 2DE or DE BC .
SIMILARITY OF TRIANGLES
If two triangles are similar, we use the symbol ||| to indicate this.
If ABC is similar to DEF then we write this as follows: ABC||| DEF
If ABC||| DEF then the following conclusions can be made:
(a) The triangles are equiangular which means that:
ˆ ˆA D ˆ ˆB E ˆ ˆC F
(b) The corresponding sides are in the same proportion which means that:
AB BC AC
DE EF DF
© Gauteng Department of Education
7
Whenever two triangles are similar we can use the following diagram to match the
corresponding angles and sides: ˆ ˆA D ˆ ˆB E ˆ ˆC F
AB BC AC
DE EF DF
THEOREM
If two triangles are equiangular then the corresponding sides of the two triangles are in
the same proportion and therefore the triangles are similar.
Proof
On AB mark off AG DE .
On AC mark off AH DF .
Join GH.
In AGH and DEF:
(1) AG DE construction
(2) ˆ ˆA D given
(3) AH DF construction
AGH DEF SAS
1ˆ ˆG E
But ˆ ˆB E given
1ˆ ˆG B
GH||BC corr 's equal
AB AC
AG AH
AB AC
DE DF ( AG DE , AH DF )
Similarly, by constructing BG and BH
on AB and BC respectively, it can be
proved that
AB BC
DE EF
AB AC BC
DE DF EF
Therefore the triangles are similar.
A
B C
D
E F
G
H
1
A
B C
D
E F
G H1
A B C D E F
© Gauteng Department of Education
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THEOREM
If the corresponding sides of two triangles
are in the same proportion, the two triangles
are similar.
Proof
Construct ABC||| GEF (as in diagram)
AB BC AC
GE EF GF
But AB BC AC
DE EF DF
AB AB
GE DE (both equal
BC
EF)
GE DE
Similarly, it can be proved that GF DF
Therefore it can be concluded that DEF GEF (SSS).
DEF||| GEF
But ABC||| GEF
ABC||| DEF
THEOREM
The perpendicular drawn from the vertex of the right angle of a right-angled triangle to
the hypotenuse, divides the triangle into two triangles that are similar to each other and
similar to the original triangle.
Proof
In ABC and DBA:
(a) 1ˆ ˆA D 90 given
(b) ˆ ˆB B common
(c) 1ˆ ˆC A sum of the 's of
ABC||| DBA
In ABC and DAC:
(a) 2ˆ ˆA D 90 given
(b) ˆ ˆC C common
(c) 2ˆB A sum of the 's of
ABC||| DAC
ABC||| DBA||| DAC
A
B C
D
E F
G
1 1
© Gauteng Department of Education
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Corollaries
2
ABC||| DBA
AB BC AC
DB BA DA
AB BD . BC
2
ABC||| DAC
AB BC AC
DA AC DC
AC CD . CB
2
DBA||| DAC
DB BA DA
DA AC DC
AD BD . DC
THEOREM (THE THEOREM OF PYTHAGORAS)
2 2 2BC AB AC
Proof
From the corollaries: 2AB BD. BC and 2AC CD. CB
2 2
2 2
2 2
2 2 2
2 2 2
AB AC BD . BC CD . CB
AB AC BC (BD CD)
AB AC BC (BC)
AB AC BC
BC AB AC
SECTION C: HOMEWORK QUESTIONS
QUESTION 1
1.1 In ABC , DE||AB, CE: EB 4:3 and
AC 28cm . Determine the length of AD. (4)
© Gauteng Department of Education
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P
Q S V R
T
U
1.2 In ACE , BG||CF and AF FE . AB 2cm, BC 3cm, CD 1cm
and DE 4cm.
1.2.1 Determine AG:GF (1)
1.2.2 Determine EG
GA (1)
1.2.3 Prove that DG||CA (4)
QUESTION 2 In PQR , PS||VT and QS:SR 2:3 .
T is a point on PR such that PT:TR 2:7 .
Prove that: QU : UT 3:1 (5)
QUESTION 3
AB is a diameter of circle ABC.
DC is a tangent at C and BD CD .
Prove that:
3.1 BDC||| BCA (5)
3.2 BD BC
CD AC (1)
QUESTION 4
© Gauteng Department of Education
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A
B CD
2
3 x
PT is a tangent to circle BAT. BA is
produced to P. TB is joined. AT is a chord.
Prove that:
BT AT
BP PT (7)
QUESTION 5
AOB is a diameter of circle centre O.
Chord AC produced meets tangent
BD at D. BC is joined.
Prove that: 2AB AC . AD (8)
QUESTION 6
TSQR is a cyclic quadrilateral.
SR||PQ and TQ bisects ˆPTR.
Prove that:
6.1 PQ is a tangent to the circle. (6)
6.2 SQ QR (5)
6.3 2QS TR .SP (10)
QUESTION 7
In ABC , A 90 , AD BC ,
BD 3, DC and AC 2x .
Calculate the length of:
7.1 DC (7)
7.2 AB (simplest surd form) (4)
© Gauteng Department of Education
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A
B
C
D
E
F
G
3BC AC
8
BC 3
AC 8
3k
8k 5k
AE :ED 4:3
4p
3p
SECTION D: SOLUTIONS FOR SECTION A
QUESTION 1
1.1 5DG 32 20
8mm mm
3GE 32 12
8mm mm
DH 5
HF 3 Line || one side of triangle
5DH 24 15
8mm mm
3HF 24 9
8mm mm
DG 20mm
GE 12mm
DH 5
HF 3
DH 15mm
HF 9mm
reason (6)
1.2 CD 14 7
DA 6 3
cm
cm
CE 21 7
EB 9 3
cm
cm
CD CE
DA EB
DE||AB Line divides sides of prop
CD 14 7
DA 6 3
cm
cm
CE 21 7
EB 9 3
cm
cm
CD CE
DA EB
DE||AB
reason (5)
1.3 DG 3
GB EF
p Line || one side of triangle; BF||GE
Now EF 3
4 8
k
p k
3
2
3EF 4
8
3EF
2
DG 3
GB
DG2
GB
DG:GB 2 :1
p
kp
k
p
p
DG 3
GB EF
p
EF 3
4 8
k
p k
3
EF2
p
3
2
DG 3
GB p
p
DG:GB 2:1 reason (6)
© Gauteng Department of Education
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QUESTION 2
2.1 In PAB and ABC:
(1) 1 2ˆ ˆA B Alt angles equal
(2) 1ˆB C Tan-chord
(3) 2ˆP A Sum of the angles of a triangle
PAB||| ABC
1 2ˆ ˆA B
1ˆB C
2ˆP A
PAB||| ABC
reason (5)
2.2 PA AB PB
AB BC AC
PA AB
AB BC
PA:AB AB:BC
PA AB PB
AB BC AC
PA:AB AB:BC (2)
2.3 PA AB
AB BC
2PA . BC AB
PA AB
AB BC
2PA . BC AB (2)
2.4 PA PB
AB AC
AP AB
BP AC
PA PB
AB AC
AP AB
BP AC (2)
QUESTION 3
In ADE and CBE:
(a) ˆ ˆE E Common
(b) 2ˆ ˆA C Ext angle of a cyclic quad
(c) 2ˆ ˆD B Sum of the angles of a triangle
ADE||| CBE
AD DE AE
CB BE CE
AD AE
CB CE
AE AD
CE BC
ˆ ˆE E
2ˆ ˆA C
2ˆ ˆD B
ADE||| CBE
AD DE AE
CB BE CE
AE AD
CE BC
reasons (7)
© Gauteng Department of Education
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QUESTION 4
4.1 In APE and CPD:
(1) 1 3ˆ ˆP P Vertically opp angles
(2) 2 2ˆ ˆA C Alt angles equal ; AE||DC
(3) 1 2ˆ ˆE D Sum of angles of
APE||| CPD
AP PE AE
CP PD CD
AP PE
PC PD
1 3ˆ ˆP P
2 2ˆ ˆA C
1 2ˆ ˆE D
APE||| CPD
AP PE
PC PD
reasons (6)
4.2 In ABP and EAP:
(1) 1 1ˆ ˆP P Common
(2) 1 2ˆB C Arc AD subtends equal angles
2 2ˆ ˆC A Alt angles equal ; AE||DC
1 2ˆB A
(3) 1 2 1ˆ ˆ ˆA A E Sum of angles of
2
2
ABP||| EAP
AB BP AP
EA AP EP
BP AP
AP EP
BP . EP AP
AP BP . PE
1 1ˆ ˆP P
1 2ˆB C
2 2ˆ ˆC A
1 2ˆB A
1 2 1ˆ ˆ ˆA A E
ABP||| EAP
AB BP AP
EA AP EP
2AP BP . PE
reasons (9)
QUESTION 5
5.1 In ABD and EDC:
(1) 1ˆ ˆA E Given
(2) 1 2ˆ ˆB D Corr angles equal
(3) 1 1ˆD C Sum of angles of
ABD||| EDC
AB BD AD
ED DC EC
AD BD
EC DC
1ˆ ˆA E
1 2ˆ ˆB D
1 1ˆD C
ABD||| EDC
AD BD
EC DC
reasons (6)
5.2 AD.DC BD.EC AD.DC BD.EC
© Gauteng Department of Education
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AD.2BC BD.BC [ DC 2BC ; BC EC ]
2AD BD
2AD BD reasons (3)
QUESTION 6
6.1 OCD||| OAC In OAC :
ˆOCA 90 , 1D 90 (diags of kite)
Perpendicular from right-angled
vertex to hypotenuse
ˆOCA 90 , 1D 90
reason (2)
6.2.1 2OC OD OA 2OC OD OA (1)
6.2.2 2CA AD AO 2CA AD AO (1)
6.2.3 2CD OD DA 2CD OD DA (1)
6.3.1 2BD OD.DA In OAB :
2OB OD.OA ˆOBA 90 , 2D 90 (diags of kite)
Perpendicular from right-angled
vertex to hypotenuse 2
2
2
2
BD OD.DA
OD.OAOB
BD AD
AOOB
2BD OD.DA
2OB OD.OA
ˆOBA 90 , 2D 90
reason
2
2
BD OD.DA
OD.OAOB
2
2
BD AD
AOOB (6)
6.3.2 2 2 2OC OD CD Pythagoras
But 2CD OD DA
2 2OC OD OD.DA
2 2 2OC OD CD
2 2OC OD OD.DA (2)
6.4 2CD OD DA 2
2 2
CD ( ).(2 )
CD 2
CD 2
CD 2.OD
x x
x
x
2CD OD DA
2CD ( ).(2 )x x
CD 2x (3)
© Gauteng Department of Education
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