Download - Geometry theorems and proofs summary
HAHS Geometry Proofs ~ 2010 1
GEOMETRY THEOREMS AND PROOFS
The policy of the HAHS Mathematics staff when teaching Geometry Proofs is to have students present a
solution in which there is a full equation showing the geometric property that is being used and a
worded reason that again identifies the geometric property that is being used.
EXAMPLE:
Find the value of x.
C
B
Ax
42
73
EQUATION REASON COMMENT
65
180115
1807342
x
x
x
(Angle sum of
180 equals ABC )
Desired level of proof to be reproduced by students
– full equation contains geometric property and
reason contains geometric property
General Notes:
(1) the word “equals” may be replaced by the symbol “=” or words such as “is”
(2) abbreviations such as “coint”, “alt”, “vert opp”, etc are not to be used – words are to be written
in full
(3) the angle symbol (), the triangle symbol (), the parallel symbol (||), the perpendicular symbol
(), etc are not to be used as substitutes for words unless used with labels such as PQR,
ABC, AB||XY, PQST
(4) If the geometric shape is not labelled then the students may introduce their own labels or refer
to the shape in general terms such as “angle sum of triangle = 180o” or “angle sum of straight
angle = 180o”
HAHS Geometry Proofs ~ 2010 2
Revolution, Straight Angles, Adjacent angles, Vertically opposite angles
The sum of angles about a point is 360o. (angles in a revolution)
Find the value of x.
P
165
60
x2x
D
C
B
A
360165602 xx (angle sum at a point P
equals 360o)
3602253 x
1353 x
45x
A right angle equals 90o.
AB is perpendicular to BC. Find the value of x.
D
CB
A
x36
9036 x (angle sum of right angle ABC
equals 90o)
54x
A straight angle equals 180o.
FMJ is a straight segment. Find the value of x.
J
I
H
G
F M
50
46 4x
2x
180504642 xx (angle sum of straight
angle FMJ equals 180o)
180966 x
846 x
14x
Three points are collinear if they form a straight angle
Given that AKB is a straight line.
Prove that the points P, K and Q are collinear.
Q
P
K
B
A
72
3x
2x
18023 xx (angle sum of straight angle AKB
equals 180o)
1805 x
36x
180
72363
723ˆ xQKP
P, K and Q are collinear (PKQ is a straight
angle) *
* PKQ equals 180o
HAHS Geometry Proofs ~ 2010 3
Vertically opposite angles are equal.
AC and DE are straight lines. Find the value of y.
y
29
D
B
E C
A
67
6729y (vertically opposite angles are equal)
38y
HAHS Geometry Proofs ~ 2010 4
Angles and Parallel Lines
Alternate angles on parallel lines are equal.
All lines are straight. Find the value of x.
>>
>>
A B
C D
E
H
F
G
x
59o
o
59x (alternate angles are equal, AB||CD)
Corresponding angles on parallel lines are equal.
All lines are straight. Find the value of x.
>>
>>
A
B
C D
E
F
G
H
137
xo
o
137x (corresponding angles are equal, AB||CD)
Cointerior angles on parallel lines are supplementary.
All lines are straight. Find the value of x.
>>
>>
A
B
C D
E
F
G
H
125
x o
o
180125x (cointerior angles are
supplementary, AB||CD)
55x
HAHS Geometry Proofs ~ 2010 5
Two lines are parallel if a pair of alternate angles are equal
Prove that AB // CD
73
73H
G
C D
AB
E
F
GHDAGH (both 73o) **
CDAB || (alternate angles are equal)
** equality of the angles involved must be clearly
indicated
Two lines are parallel if a pair of corresponding angles are equal
Prove that AB // CD
65
65H
G
C D
AB
E
F
EGB = GHD (both 65o) **
CDAB || (corresponding angles are equal)
** equality of the angles involved must be clearly
indicated
Two lines are parallel if a pair of cointerior angles are supplementary
Prove that PR // KM
56
124
L
Q
K M
PR
X
Y
RQL + QLM = 124o + 56
o **
= 180o
KMPR || (cointerior angles are
supplementary)
* RQL + QLR = 180o
** supplementary nature of the angles involved
must be clearly indicated
HAHS Geometry Proofs ~ 2010 6
Angles in Polygons
The angle sum of a triangle is 180o.
Find the value of x.
A
B
C
x34
67
o
o
o
1803467 x (angle sum of ABC equals
180o)
180101x
79x
The exterior angle of a triangle equals the sum of the opposite (or remote) interior angles.
Find the value of x.
A DC
B
x
47
68oo
o
4768x (exterior angle of ABC equals sum
of the two opposite interior angles)
115x
* exterior angle of ABC equals sum of remote
interior angles
The angle sum of the exterior angles of a triangle is 360o.
Find the value of x.
A C
B
x
157
128o
o
o
360128157 x (sum of exterior angles of
ABC equals 360o)
360285x
75x
The angles opposite equal sides of a triangle are equal. (converse is true)
Find the value of x.
||
=
A B
C
54
xo
o
54x (equal angles are opposite equal sides in
ABC ) *
* base angles of isosceles ABC are equal
HAHS Geometry Proofs ~ 2010 7
The sides opposite equal angles of a triangle are equal (converse is true).
Find the value of x.
12
15
x
A
B C
65o
65o
15x (equal sides are opposite equal angles in
ABC )
All angles at the vertices of an equilateral triangle are 60o.
ABC is equilateral. EC and DB are angle
bisectors and meet at P. Find the size of CPB.
BC
A
PD E
ACB = 60o (all angles of an equilateral triangle
are 60o)
similarly ABC = 60o
ECB = 30o (EC bisects ACB)
similarly DBC = 30o
CPB + 60o = 180
o (angle sum of PCB equals
180o)
CPB = 120o
The angle sum of a quadrilateral is 360o.
Find the value of x.
A
B
C
D
o
o
o
ox
3x
130
70
3602004 x (angle sum of quadrilateral ABCD
equals 360o)
40
1604
x
x
HAHS Geometry Proofs ~ 2010 8
The angle sum of a n-sided polygon is 180(n – 2)o or (2n – 4) right angles.
Find the value of x.
106
87
x
165
92
B
C
D
E
A
Angle sum of a pentagon = 3 180o
= 540o
x + 450 = 540 (angle sum of pentagon equals
540o)
x = 90
The angle at each vertex of a regular n-sided polygon is o
2180
n
n.
Find the size of each interior angle of a regular
hexagon
120
6
4180 size Angle
The angle sum of the exterior angles of a n-sided polygon is 360o.
Find the size of each interior angle of a regular
decagon.
Sum of exterior angles = 360o
Exterior angles =
o
10
360
= 36o
Interior angles = 144o (angle sum of straight angle
equals 180o)
HAHS Geometry Proofs ~ 2010 9
Similar Triangles
Two triangles are similar if two angles of one triangle are equal to two angles of the other triangle.
Prove that ABC and DCA are similar.
**
A
B
C
D
In ABC and DCA
ABC = ACD (given)
BAC = ADC (given)
ABC ||| DCA (equiangular) *
* The abbreviations AA or AAA are not to be
accepted
Two triangles are similar if the ratio of two pairs sides are equal and the angles included by these
sides are equal.
Prove that ABC and ACD are similar.
36
16
24
**
A
B
C
D
In ABC and ACD
BCA = ACD (given)
2
3
24
36
AC
BC
2
3
16
24
DC
AC
BCA ||| ACD (sides about equal angles are in
the same ratio) *
* sides about equal angles are in proportion
Two triangles are similar if the ratio of the three pairs of sides are equal.
Prove that ABC and ACD are similar.
A
B C
D
12
16
24
18
32
In ABC and ACD
3
4
12
16
CD
AB
3
4
24
32
AC
BC
3
4
18
24
AD
AC
ABC ||| DCA (three pairs of sides in the same
ratio) *
* three pairs of sides in proportion
HAHS Geometry Proofs ~ 2010 10
Example:
Given that PQAB // , find the value of x.
9 cm
Q
P
CB
A
x cm
12 cm8 cm
In ABC and PQC
PQCABC (corresponding angles are equal
as PQAB // )
PCQACB (common)
PQCABC ||| (equiangular)
12
20
9
x (corresponding sides in similar triangles
are in the same ratio) *
12
209x
15x
* corresponding sides in similar triangles are in
proportion
HAHS Geometry Proofs ~ 2010 11
Congruent Triangles
Two triangles are congruent if three sides of one triangle are equal to three sides of the other
triangle.
Given that AC = BD and AB = CD.
Prove that CAB BDC.
12
8
12
8
A
B
C
D
In CAB and BDC.
AC = BD (both 8) or (given) or (data)
AB = CD (both 12) or (given) or (data)
CB = CB (common) or CB is common
CAB BDC (SSS)
or
In CAB and BDC.
AC = BD = 8
AB = CD = 12
CB = CB (common) or CB is common
CAB BDC (SSS)
Two triangles are congruent if two sides of one triangle are equal to two sides of the other triangle
and the angles included by these sides are equal.
Given that AC = BD and CAB = DBA.
Prove that CAB DBA.
= =
A B
CD
In CAB and DBA
AB = AB (common) or AB is common
AC = BD (given)
CAB = DBA (given)
CAB DBA (SAS)
Two triangles are congruent if two angles of one triangle are equal to two angles of the other
triangle and one pair of corresponding sides are equal.
Given that AB = CD and EAB = ECD.
Prove that ABE CDE.
= =
A
B
C
D
E
* *
In ABE and CDE.
AB = CD (given)
EAB = ECD (given)
AEB = CED (vertically opposite angles are
equal)
ABE CDE (AAS)
HAHS Geometry Proofs ~ 2010 12
Two right- angled triangles are congruent if their hypotenuses are equal and a pair of sides are also
equal.
Given that CD = AD. Prove that ABD CBD.
=
=
A
BD
C
In ABD and CBD
BCD = BAD (both 90o)
CD = AD (given)
DB = DB (common)
ABD CBD (RHS)
HAHS Geometry Proofs ~ 2010 13
Intercepts and Parallels
An interval joining the midpoints of the sides of a triangle is parallel to the third side and half its
length.
E and F are midpoints of AB and AC.
G and H are midpoints of FB and FC.
Prove that EF = GH.
B C
A
E F
G H
EF=½BC (interval joining midpoints of sides of
ABC is half the length 3rd side)
Similarly in BFC , GH=½BC
EF = GH
(Note: It can also be proven that EF and GH are
parallel)
An interval parallel to a side of a triangle divides the other sides in the same ratio. (converse is true)
Find the value of x.
>
>
B C
A
I J
x
15 9
20
15
20
9
x (interval parallel to side of ABC divides
other sides in same ratio)
x = 12
Parallel lines preserve the ratio of intercepts on transversals. (converse is not true)
Find the value of x.
>
>
>
x
24
32
18
24
18
32
x (parallel lines preserve the ratios of
intercepts on transversals) *
x = 24
* intercepts on parallel lines are in the same ratio
* intercepts on parallel lines are in proportion
HAHS Geometry Proofs ~ 2010 14
Pythagoras’ Theorem
Pythagoras’ Theorem: The square on the hypotenuse equals the sum of the squares on the other two
sides in a right angled triangle.
Find the value of x.
12
15
x
222 1512 x (Pythagoras’ Theorem)
9
81
1442252
x
x
or
9x (3,4,5 Pythagorean Triad)
A triangle is right-angled if the square on the hypotenuse equals the sum of the squares on the other
two sides (converse of Pythagoras’ Theorem)
Prove that ABC is right-angled
8 cm
10 cm6 cm
A C
B
222
2222
22
100
6436
86
100
10
BCACAB
ACAB
BC
ABC is right-angled (Converse of Pythagoras’
theorem)
HAHS Geometry Proofs ~ 2010 15
Quadrilateral Properties
Trapezium
One pair of sides of a trapezium are parallel
The non-parallel sides of an isosceles trapezium are equal
Parallelogram
The opposite sides of a parallelogram are parallel
The opposite sides of a parallelogram are equal
The opposite angles of a parallelogram are equal
The diagonals of a parallelogram bisect each other
Adjacent angles are supplementary
A parallelogram has point symmetry
Kite
Two pairs of adjacent sides of a kite are equal
One diagonal of a kite bisects the other diagonal
One diagonal of a kite bisects the opposite angles
The diagonals of a kite are perpendicular
A kite has one axis of symmetry
Rhombus
The opposite sides of a rhombus are parallel
All sides of a rhombus are equal
The opposite angles of a rhombus are equal
The diagonals of a rhombus bisect the opposite angles
The diagonals of a rhombus bisect each other
The diagonals of a rhombus are perpendicular
A rhombus has two axes of symmetry
A rhombus has point symmetry
Rectangle
The opposite sides of a rectangle are parallel
The opposite sides of a rectangle are equal
All angles at the vertices of a rectangle are 90o
The diagonals of a rectangle are equal
The diagonals of a rectangle bisect each other
A rectangle has two axes of symmetry
A rectangle has point symmetry
Square
Opposite sides of a square are parallel
All sides of a square are equal
All angles at the vertices of a square are 90o
The diagonals of a square are equal
The diagonals of a square bisect the opposite angles
The diagonals of a square bisect each other
The diagonals of a square are perpendicular
A square has four axes of symmetry
A square has point symmetry
HAHS Geometry Proofs ~ 2010 16
Sufficiency conditions for Quadrilaterals
Sufficiency conditions for parallelograms
A quadrilateral is a parallelogram if
both pairs of opposite sides are parallel or
both pairs of opposite sides are equal or
both pairs of opposite angles are equal or
the diagonals bisect each other or
one pair of sides are equal and parallel
Sufficiency conditions for rhombuses
A quadrilateral is a rhombus if
all sides are equal or
the diagonals bisect each other at right angles
or
the diagonals bisect each vertex angle
Sufficiency conditions for rectangles
A quadrilateral is a rectangle if
all four angles are equal or
the diagonals are equal and bisect each other
Sufficiency condition for squares
A quadrilateral is a square if
the diagonals are equal and bisect each other at
right angles
Sufficiency condition for kites
A quadrilateral is a kite if
the diagonals are perpendicular and one is bisected
by the other
HAHS Geometry Proofs ~ 2010 17
Circles and Chords or Arcs
Equal chords subtend equal arcs on a circle. (converse is true)
Equal arcs subtend equal chords on a circle. (converse is true)
Equal chords subtend equal angles at the centre of a circle. (converse is true)
AB = EF. Find the value of x.
x68
O
E
A
F
B
x = 68 (equal chords subtend equal angles at the
centre)
Equal arcs subtend equal angles at the centre of a circle. (converse is true)
arc AB = arc EF. Find the value of x.
x68
O
E
A
F
B
x = 68 (equal arcs subtend equal angles at the
centre)
Equal angles at the centre of a circle subtend equal chords. (converse is true)
Chord EF = 16cm, find the length of chord AB.
O
F
E
B
A
7575
AB = 16 cm (equal angles at the centre subtend
equal chords)
HAHS Geometry Proofs ~ 2010 18
Equal angles at the centre of a circle subtend equal arcs. (converse is true)
arc EF = 16cm, find the length of arc AB.
16 cm
O
F
E
B
A
7575
arc AB = 16 cm (equal angles at the centre subtend
equal arcs)
A line through the centre of a circle perpendicular to a chord bisects the chord. (converse is true)
O is the centre of the circle. Find the length of AP.
8 cm
O
B
A
P
AP = 8 cm (interval through center perpendicular to
chord AB bisects the chord)
A line through the centre of a circle that bisects a chord is perpendicular to the chord. (converse is true)
Find the size of OEB.
6 cm
6 cm
E
O
B
C
chord thelar toperpendicu is chord
bisecting centre through interval90
BCOEB
NOTE:
It can be proven that the perpendicular bisector of a chord passes through the centre of the circle.
HAHS Geometry Proofs ~ 2010 19
Chords equidistant from the centre of a circle are equal. (converse is true)
Find the length of XY.
5cm
=
=O
B
AP
Y
XQ
AB = 10 cm (interval through centre perpendicular
to chord AB bisects the chord)
XY = 10 cm (chords equidistant from the centre of
a circle are equal)
Equal chords are equidistant from the centre of a circle. (converse is true)
Find the length of OL.
7
75
7
7
LM
OH
I
G
F
IH = FG = 14
OL = 10 (equal chords are the equidistant from the
centre)
HAHS Geometry Proofs ~ 2010 20
Angles in Circles
The angle at the centre of a circle is twice the angle at the circumference standing on the same arc.
The angle at the circumference of a circle is half the angle at the centre standing on the same arc.
(i) Find the value of y.
O
B
A
C
54
y
(ii) Find the value of x.
O
B
A
C
94
x
(i) y = 108 (angle at centre equals twice angle
circumference standing on arc AB)
Note: use arc AB and not chord AB – the
statement is not necessarily true for
chords
(ii) x = 47 (angle at circumference equals half
angle at centre standing on arc AB)
Angles at the circumference standing on the same arc are equal
or
Angles at the circumference in the same segment are equal. (converse is true)
Find the value of x.
S
O
P
R
Q
41
x
x = 41 (angles at the circumference on the same
arc PQ are equal)
(Note: use arc PQ and not chord PQ – the
statement is not necessarily true for chords)
or
x = 41 (angles at the circumference in the same
segment equal)
HAHS Geometry Proofs ~ 2010 21
Equal arcs subtend equal angles at the circumference. (converse is true)
arc AB = arc CD. Find the value of x.
x
37
D
A
B
C
FE
x = 37 (Equal arcs subtend equal angles at the
circumference)
Note: the statement is not necessarily true for
equal chords
Equal angles at the circumference subtend equal arcs.
Find the length of arc PQ.
8 cm
25
25N
Q
Y
X
P
M
PQ = 8 cm (Equal angles at the circumference
subtend equal arcs)
The angle at the circumference in a semi-circle is 90o.
AB is a diameter. Find the value of x.
38x
A O B
P
90ˆBPA (angle at circumference in semi-circle
equals 90o)
x + 128 = 180 (angle sum of APB equals 180o)
x = 52
A right angle at the circumference subtends a diameter
If 90ˆBCA then AB is a diameter.
BA
C
HAHS Geometry Proofs ~ 2010 22
A radius (diameter) of a circle is perpendicular to the tangent at their point of contact
STU is a tangent at T. Find the size of TOU.
26
O
T
U
S
OTU = 90o (radius is perpendicular to tangent at
point of contact)
TOU + 116o = 180
o (angle sum of OUT equals
180o)
TOU = 64o
The angle between a tangent and a chord equals the angle at the circumference in the alternate
segment.
Find the size of RTN.
93
T
N
M
R
S
RTN = 93o (angle between tangent and chord
equals angle at circumference in
alternate segment)
* Alternate segment theorem
HAHS Geometry Proofs ~ 2010 23
Cyclic Quadrilaterals
The opposite angles of a cyclic quadrilateral are supplementary. (converse is true)
Find the value of x.
C
D
B
A
87
x o
o
x + 87 = 180 (opposite angles of cyclic
quadrilateral ABCD are
supplementary)
x = 93
* opposite angles of a cyclic quadrilateral are
supplementary
The exterior angle of a cyclic quadrilateral equals the opposite (or remote) interior angle. (converse
is true)
Find the size of ADE.
DCE
B
A
o112
ADE = 112o (exterior angle of cyclic
quadrilateral ABCD equals
opposite interior angle)
or
ADE = 112o (exterior angle of cyclic
quadrilateral ABCD equals
remote interior angle)
HAHS Geometry Proofs ~ 2010 24
Intercept Theorems
The product of the intercepts on intersecting chords are equal. (converse is true)
Find the value of x.
x
12
QA
P
B
18
8
x 8 = 12 18 (product of intercepts on
intersecting chords are equal)
x = 27
The product of the intercepts on intersecting secants are equal.
Find the value of x.
x
A
P Q
B
T
93
12
121599 x (product of intercepts on
intersecting secants are equal)
9x + 81 = 180
9x = 99
x = 11
The square of the intercept on tangent to a circle equals the product of the intercepts on the secant.
Find the value of x.
x
12
T
B
A
P4
4162 x (square of intercept on tangent to
circle equals product of intercepts
on secant)
x2 = 64
x = 8 (length > 0)
HAHS Geometry Proofs ~ 2010 25
Intercepts on tangents drawn from a point to a circle are equal.
Find the value of x.
x
35
x = 35 (intercepts on tangents
from a point to a circle
are equal)
The line joining the centers of two circles passes through their point of contact
HAHS Geometry Proofs ~ 2010 26
Converses of Cyclic Quadrilateral theorems
If the opposite angles of a quadrilateral are supplementary then the quadrilateral is cyclic.
XA and YB are altitudes of XYZ. Prove that AZBP
is a cyclic quadrilateral.
X
Y
Z
A
B
P
YBZ = 90o (YB is an altitude)
XAZ = 90o (XA is an altitude)
PBZ + PAZ = 180o
AZBP is cyclic (opposite angles are
supplementary)
If the exterior angle of a quadrilateral equals the opposite interior angle then the quadrilateral is
cyclic.
Prove that ABCD is a cyclic quadrilateral.
87
87
AB
C
D
T
o
o
DAB = TCB (both 87o)
ABCD is a cyclic (exterior angle equals opposite
interior angle)
If a side of a quadrilateral subtends equal angles at the other two vertices then the quadrilateral is
cyclic.
OR
If an interval subtends equal angles on the same side at two points then the ends of the interval and
the two points are concyclic.
XA and YB are altitudes of XYZ. Prove that XBAY
are the vertices of a cyclic quadrilateral.
X
Y
Z
A
B
P
XBY = 90o (YB is an altitude)
XAY = 90o (XA is an altitude)
XBA = XAY = 90o
XBAY is cyclic (XY subtends equal angles on
the same side at A and B)
HAHS Geometry Proofs ~ 2010 27
If the product of the intercepts on intersecting intervals are equal then the endpoints of the intervals
are concyclic.
Prove that points A, C, B and D are concyclic.
A
B
C
D
F
4
69
6
36 FCDFFBAF
A, C, B and D are concyclic (product of
intercepts are equal)