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”


Revolution, Straight Angles, Adjacent angles, Vertically opposite angles

The sum of angles about a point is 360o. (angles in a revolution)


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


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


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.


>>

>>

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.


>>

>>

A

B

C D

E

F

G

H

125

x o

o

180125x (cointerior angles are

supplementary, AB||CD)

55x


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


Angles in Polygons

The angle sum of a triangle is 180o.


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.


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.


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)


||

=

A B

C

54

xo

o

54x (equal angles are opposite equal sides in

ABC ) *

* base angles of isosceles ABC are equal


The sides opposite equal angles of a triangle are equal (converse is true).


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.


A

B

C

D

o

o

o

ox

3x

130

70

3602004 x (angle sum of quadrilateral ABCD

equals 360o)

40

1604

x

x


The angle sum of a n-sided polygon is 180(n – 2)o or (2n – 4) right angles.


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)


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


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


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)


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)


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)


>

>

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)


>

>

>

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


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.


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)


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


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


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)


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.


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)


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)


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)


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


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


Cyclic Quadrilaterals

The opposite angles of a cyclic quadrilateral are supplementary. (converse is true)


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)


Intercept Theorems

The product of the intercepts on intersecting chords are equal. (converse is true)


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.


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.


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)


Intercepts on tangents drawn from a point to a circle are equal.


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


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)


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)

geometry theorems and proofs summary

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