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Specialist Mathematics Essentials
Geometric Proof
Vector method of proof
Vectors can often be used to establish geometric properties, especially involving parallelism,
perpendicularity and properties of intersections.
Summary of rules
Property
Parallelism
Collinearity
Midpoint
Algebraic forms
Geometric form
a kb, k z 0
a k b
JJJG
JJJJG
AB k BC
Ÿ A, B and C are collinear
JJJG
Ÿ B divides AC in ratio k : l
JJJG
JJJG
If OA a, OB b
JJJJG 1
1
a b
the OM
2 2
Perpendicularity
and
Scalar Product
a ˜ b 0 œ a and b are perpendicular
Non-parallel vectors
If k1a k2 b l1a l2 b and a and b
are not parallel, then k1 l1 and k 2 l2
Refer to example.
Geometric Proof
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Example
Prove, by vector means, that ‘the angle subtended by a diameter is 90q ’.
Let the centre of the circle be O, where AB is the diameter.
JJJG JJJG
JJJG
Let OA BO a and OC c
JJJG
? AC a c
JJJG
BC a c
G JJJG JJJ
? AC ˜ BC a c ˜ a c a ˜ a a ˜ c c ˜ a c c
2 2 a c , since a ˜ c c ˜ a
but a c radius of circle
JJJ
G
JJJG
? AC ˜ BC 0
JJJG
JJJG
? AC is perpendicular to BC (for all positions of C)
?
angle subtended by a diameter is 900.
Example
JJJG
O divides AD in the ratio 2:1, and B is the
midpoint of OC.
JJJG
JJJG
OA 2a, and OB b .
DB produced meets AC at M.
Determine the position vector of M
terms of a and b .
1)
JJJJG JJJG JJJJG
OM OA AM
JJJG
JJJG
OA s AC
2)
2a s 2a 2b 2 2s a 2s b
From 1) and 2):
JJJJG JJJG JJJJG
OM OD DM
JJJG JJJG
OD t DB
a t ( a b)
1 t a t b
2 2 s a 2 s b 1 t a t b
? 2 2s
? 2 2s
? 4s 3
1 t and 2s t
1 2 s
3
3
, t
4
2
3 JJJG JJJJG 3 JJJG
AC , DM
DB and
4
2
? s
Thus
JJJJG
AM
JJJJG
OM
1
3
a b.
2 2
JJJJG
i.e. OM in
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Specialist Mathematics Essentials
Exercise 1
A trapezium OABC has a vertex O at the origin.
JJJG
JJJG
JJJG
OA a , OC c and AB k c for some number k ! 0 .
JJJG
JJJG
M and N are the midpoints of OA and BC respectively.
JJJG
a)
Show that BC 1 k c a .
JJJJG
JJJG
k 1
times its
b)
Show that MN is parallel to OC and is
2
length.
Exercise 2
Use the vector diagram given, to prove that ‘the sum of the squares of the lengths of the diagonals of a
parallelogram is equal to the sum of the squares of the lengths of its sides’.
JJJG 2
Hint: Use the fact that OB
JJJG JJJG
OB ˜ OB
Geometric Proof
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Exercise 3
AM and CN are medians of 'ABC (i.e. M and N are
midpoints of sides BC and BA respectively).
X is their point of intersection.
JJJG
JJJJG
Let AN a and BM b .
JJJG
JJJJG
JJJG
JJJG
Let AX k AM and CX l CN .
a)
b)
c)
JJJG
By using the fact that AC
JJJG JJJG
AX XC
JJJG JJJG
AX CX , show that k
Suppose that P is the midpoint of AC.
JJJG
JJJG
JJJG
By expressing BX and BP in terms of a and b , show that BX
Which property of a triangle have you just proved?
l
2
.
3
2 JJJG
BP .
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Exercise 4
In the diagram above, OACB is a parallelogram with angle AOB of measure T . Let
JJJG
JJJG
JJJG
OA p and OB q, where q 2 p . Let AD k q, 0 d k d 1 , where D is a point on AC such that angle
ODB measures 900.
a)
b)
c)
d)
2
Show that p ˜ q 2 p cosT .
JJJG JJJG Find OD and DB in terms of p, q and k .
2
2
0 and hence deduce that
Using b), show that 1 2k p ˜ q 1 2k p
1
1
k
or k
cosT .
2
2
1
Describe the positions of D for which k
, k 0 and k 1 .
2
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Specialist Mathematics Essentials
Geometry of circles
Circle geometry is part of Deductive (or Euclidean) geometry.
Deductive geometry uses highly developed logical reasoning, together with special, known results called
theorems, to prove that certain observations about geometrical figures are indeed true.
Angle properties
Name of theorem
Angle at the centre
Statement
The angle subtended by
an arc (or chord) of a
circle at the centre is
twice the angle subtended
by the same arc (or
chord) at the
circumference.
Angles subtended by
the same arc
or
Angles subtended by the
same arc of a circle are
equal.
Angles in the same
segment
Angle in a semicircle.
Angles subtended by a
diameter (angles in a
semi-circle) are right
angles.
Diagrams
Geometric Proof
Tangent Properties
Name of theorem
Statement
Radius-tangent
The radius drawn to the
point of contact of a
tangent is
perpendicular to the
tangent.
Tangents from an
external point
Tangents drawn from
an external point are
equal in length.
Angle between a
tangent and chord
The angle between a
tangent and a chord at
the point of contact is
equal to the angle
subtended by the chord
in the alternate
segment.
Cyclic quadrilateral properties
Opposite angles of a cyclic
quadrilateral
The opposite angles of a cyclic
quadrilateral are supplementary.
Exterior angle of a cyclic
quadrilateral
The exterior angle of a cyclic
quadrilateral is equal to the
interior opposite angle.
Diagrams
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Specialist Mathematics Essentials
Tests for concyclic points (4 points)
When the 4 points are joined to form a convex quadrilateral and one pair of opposite angles are
supplementary.
When two points (defining a line) subtend equal angles at the other two points on the same side of the
line.
Example
AB and CD are two chords of a circle, which when
produced meet at point X.
a)
Show that triangles ACX and DBX are
similar.
b)
Hence prove that AX .BX CX .DX .
c)
Use the result in b) to find the length of AB
in the following diagram:
a)
‘X ‘CXA ‘DXB (common angle)
‘DBX ‘ACX (exterior angle of cyclic quad. ABDX )
‘BDX ‘CAX (exterior angle of cyclic quad. ABDX )
? 's ACX and DBX are similar (equiangular)
b)
Similar 's Ÿ sides in proportion
AX CX
?
DX BX
? AX .BX CX .DX .
c)
Using result in b):
AB 2 2
5u 3
? 2 AB 4 15
? 2 AB 11
11
? AB
5˜5
2
Geometric Proof
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Example
AB and AC are equal chords of a circle. P and Q are two points on the chord BC, and AP and AQ, when
extended, meet the circle again at points R and S.
Prove that PQSR is a cyclic quadrilateral.
Hint: Join SC. Let ‘SAC T , ‘ABC D . Find ‘ACS in terms of T and D .
Let ‘ABC D .
'ABC is isosceles AB AC ? ‘ABC ‘ACB D (base angles equal)
‘ABC and ‘ASC are subtended by the same chord AC.
? ‘ABC ‘ASC D (angles subtended by same chord are equal).
In ' ASC : ‘ACS 180 T D (sum of angles in ' 180q )
ARSC is a cyclic quadrilateral.
? ‘ARS 180 ‘ACS
(opposite angles of a cyclic quadrilateral are supplementary)
180 ª¬180 T D º¼
T D
In ' AQC : ‘AQS 180 T D ? ‘PQS 180 T D (sum of angles in ' =180q)
‘PQS
‘AQC , vertically opposite Consider quadrilateral PQSR:
‘PQS ‘PRS 180 T D T D
180q
? PQSR is a cyclic quadrilateral
(opposite angles supplementary).
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Specialist Mathematics Essentials
Example
AB and CD are two chords of a circle, intersecting at
the point P. Q is a point such that ‘QAB and
‘QCD are right angles.
QP is produced to meet BD at the point T.
a)
Prove that APCQ is a cyclic quadrilateral.
b)
Prove that ‘QCA ‘BPT .
c)
Hence, or otherwise, prove that QT is
perpendicular to BD.
d)
i)
Prove that the triangle APQ and the
triangle TPB are similar.
ii)
If AQ 12 cm, PQ 13 cm, and
BT 16 cm, find QT .
a)
‘QCP ‘QAP 90q 90q
?
b)
c)
180q
APCQ is a cyclic quadrilateral (opposite angles supplementary).
Let ‘QCA D
? ‘QCA ‘QPA D
(angles in same segment)
? ‘QPA ‘BPT D
i.e. ‘QCA ‘BPT
(vertically opposite)
‘ACP ‘ABT E
But D E 90q
(angles in same segment)
(angle QCP)
?
in 'PBT , ‘BTP 180q D E 180q 90q
90q
? QT A BD .
d)
i)
' s APQ and TPB are right-angled and have equal D angles and equal E angles.
? triangles are equiangular and hence similar.
ii)
Now,
x
5
(sides in proportion)
16 12
5
? x
u 16
12
2
? x 6
3
2
? QT 13 6
3
2
19 cm
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Geometric Proof
Exercise 5
Three intersecting circles are shown below. A, B, C, D and E, F, G, H are collinear points lying on the
circles, as shown in the figure.
Prove that
a)
AE is parallel to CG.
b)
ADHE is a cyclic quadrilateral.
Exercise 6
The points R, S and T lie on the circumference of a circle with
centre O, as shown below.
The tangent T and the chord SR produced meet at the point Q.
The angle ‘STR 30q .
a)
b)
Show that the length of SR equals the radius of the
circle.
If ‘TQR 40q, find ‘RTQ.
Exercise 7
AB is a chord of a circle which, when produced, meets the tangent at T at the point X.
a)
Prove that triangles AXT and TXB are similar.
b)
Hence show that AX .BX TX 2 .
Exercise 8
In the diagram shown, AB is the diameter and AC is a chord of a circle. AP
is a tangent at A. PQ is parallel to AC and meets BC in Q.
a)
Prove that the points P, A, Q and B are concyclic.
b)
Calculate the size of ‘APB , given that ‘CAQ 35q .
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Specialist Mathematics Essentials
Exercise 9
XY is any chord of circle. XY is produced to T and TP is a tangent to
the circle. The angle bisector of angle PTX cuts PY at L and PX at M.
Show that PM PL .
Exercise 10
ABC is a triangle inscribed in a circle. PA is a tangent to the
circle at A. X and Y are points on sides AB and AC respectively,
such that XY is parallel to AP.
Prove that BXYC is a cyclic quadrilateral.
Exercise 11
A, B and C are three points on a circle. D and E lie on AB and AC
respectively, such that DE is parallel to BC. If CD produced and BE
produced meet the circle at X and Y respectively, show that XYED is a
cyclic quadrilateral.
Exercise 12
AB is a diameter of a circle. PQ is a chord perpendicular to AB, and
meets AB at a point R. S is a point on AR, and QS produced meets the
circle at T.
a)
Prove that triangles PSR and QSR are congruent.
b)
Hence prove that AP bisects angle TPS.
Hint: Join AQ.