Geometry theorems and proofs summary
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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.
B 73
42 A
x C
EQUATION x 42 73 180
x 115 180 x 65
REASON (Angle sum of ABC equals 180 )
COMMENT 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
1
Revolution, Straight Angles, Adjacent angles, Vertically opposite angles The sum of angles about a point is 360o. (angles in a revolution)
2x x 60 165 360 (angle sum at a point P equals 360o) 3x 225 360 3x 135 x 45
Find the value of x. C B 60 x
2x P 165
A
D
A right angle equals 90o. AB is perpendicular to BC. Find the value of x. A
x 36 90 (angle sum of right angle ABC equals 90o) x 54
D x
36 C
B
A straight angle equals 180o. FMJ is a straight segment. Find the value of x. H I G
2x 4x 46 50 180 (angle sum of straight angle FMJ equals 180o) 6 x 96 180 6x 84 x 14
4x
46 2x
50 M
F
J
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. A P 3x 72
Q
HAHS Geometry Proofs ~ 2010
K
2x B
3x 2x 180 (angle sum of straight angle AKB equals 180o) 5x 180 x 36 PKˆ Q 3x 72 3 36 72 180 P, K and Q are collinear (PKQ is a straight angle) * o * PKQ equals 180
2
Vertically opposite angles are equal. AC and DE are straight lines. Find the value of y. A
y 29 67 (vertically opposite angles are equal) y 38
D
67
E
HAHS Geometry Proofs ~ 2010
29 y
B
C
3
Angles and Parallel Lines Alternate angles on parallel lines are equal.
x 59 (alternate angles are equal, AB||CD)
All lines are straight. Find the value of x. E C x
A
G
o
D
>>
F
o
59
B
>>
H
Corresponding angles on parallel lines are equal. All lines are straight. Find the value of x.
x 137 (corresponding angles are equal, AB||CD)
E x A
o
>>
G
o
137 C
B
>>
H
D
F
Cointerior angles on parallel lines are supplementary. All lines are straight. Find the value of x. E
A
G
x
>>
o
125 C
x 125 180 (cointerior angles are supplementary, AB||CD) x 55
B
o
H
>>
D
F
HAHS Geometry Proofs ~ 2010
4
Two lines are parallel if a pair of alternate angles are equal Prove that AB // CD
AGH GHD (both 73o) ** AB || CD (alternate angles are equal)
E A
G
B
73
** equality of the angles involved must be clearly indicated
H 73
C
D
F
Two lines are parallel if a pair of corresponding angles are equal EGB = GHD (both 65o) ** AB || CD (corresponding angles are equal)
Prove that AB // CD E A
G 65
B
** equality of the angles involved must be clearly indicated H 65
C
D
F
Two lines are parallel if a pair of cointerior angles are supplementary RQL + QLM = 124o + 56o ** = 180o
Prove that PR // KM X R
Q
P
124
PR || KM (cointerior angles are supplementary) * RQL + QLR = 180o
L
K
56
M
** supplementary nature of the angles involved must be clearly indicated Y
HAHS Geometry Proofs ~ 2010
5
Angles in Polygons The angle sum of a triangle is 180o.
x 67 34 180 (angle sum of ABC equals 180o) x 101 180 x 79
Find the value of x.
A o
67
x
o o
B
34
C The exterior angle of a triangle equals the sum of the opposite (or remote) interior angles.
x 68 47 (exterior angle of ABC equals sum of the two opposite interior angles) x 115
Find the value of x. B o
47
* exterior angle of ABC equals sum of remote interior angles
o
xo
68
A
D
C
The angle sum of the exterior angles of a triangle is 360o.
x 157 128 360 (sum of exterior angles of ABC equals 360o) x 285 360 x 75
Find the value of x.
B o
157
o
128
A
xo
C
The angles opposite equal sides of a triangle are equal. (converse is true)
x 54 (equal angles are opposite equal sides in ABC ) *
Find the value of x. C 54
o
=
* base angles of isosceles ABC are equal ||
A HAHS Geometry Proofs ~ 2010
xo B 6
The sides opposite equal angles of a triangle are equal (converse is true).
x 15 (equal sides are opposite equal angles in ABC )
Find the value of x. A o
65
12
x
o
65
B
C
15
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. A
E
D
ACB = 60o (all angles of an equilateral triangle are 60o) similarly ABC = 60o ECB = 30o (EC bisects ACB) similarly DBC = 30o CPB + 60o = 180o (angle sum of PCB equals 180o) CPB = 120o
P
C
B
The angle sum of a quadrilateral is 360o.
4x 200 360 (angle sum of quadrilateral ABCD equals 360o) 4 x 160 x 40
Find the value of x. A o
130
3x o
D
xo B
o
70
C
HAHS Geometry Proofs ~ 2010
7
The angle sum of a n-sided polygon is 180(n – 2)o or (2n – 4) right angles. Find the value of x. B 87
C
106
A
Angle sum of a pentagon = 3 180o = 540o x + 450 = 540 (angle sum of pentagon equals 540o) x = 90
92 165
D x
E
180n 2 The angle at each vertex of a regular n-sided polygon is . n o
Find the size of each interior angle of a regular hexagon
180 4 Angle size 6 120
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.
HAHS Geometry Proofs ~ 2010
Sum of exterior angles = 360o o
360 Exterior angles = 10 = 36o Interior angles = 144o (angle sum of straight angle equals 180o)
8
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.
In ABC and DCA ABC = ACD (given) BAC = ADC (given) ABC ||| DCA (equiangular) *
D
A
* The abbreviations AA or AAA are not to be accepted B
*
*
C
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.
A B D
24
In ABC and ACD BCA = ACD (given) BC 36 3 AC 24 2 AC 24 3 DC 16 2 BCA ||| ACD (sides about equal angles are in the same ratio) *
36 * sides about equal angles are in proportion
16
* * C
Two triangles are similar if the ratio of the three pairs of sides are equal. Prove that ABC and ACD are similar. A
18
D 16 24
B
32
12
C
In ABC and ACD AB 16 4 CD 12 3 BC 32 4 AC 24 3 AC 24 4 AD 18 3 ABC ||| DCA (three pairs of sides in the same ratio) * * three pairs of sides in proportion
HAHS Geometry Proofs ~ 2010
9
Example: Given that AB // PQ , find the value of x.
A P
x cm 9 cm B
8 cm Q
12 cm
C
In ABC and PQC ABC PQC (corresponding angles are equal as AB // PQ ) ACB PCQ (common) ABC ||| PQC (equiangular) x 20 (corresponding sides in similar triangles 9 12 are in the same ratio) * 20 x 9 12 x 15
* corresponding sides in similar triangles are in proportion
HAHS Geometry Proofs ~ 2010
10
Congruent Triangles Two triangles are congruent if three sides of one triangle are equal to three sides of the other triangle. 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)
Given that AC = BD and AB = CD. Prove that CAB BDC. 8
A
C 12
or
12 B
D
8
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. C
D
=
=
A
B
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. C
A
*
* = B
HAHS Geometry Proofs ~ 2010
E
In ABE and CDE. AB = CD (given) EAB = ECD (given) AEB = CED (vertically opposite angles are equal) ABE CDE (AAS)
= D
11
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. C
=
In ABD and CBD BCD = BAD (both 90o) CD = AD (given) DB = DB (common) ABD CBD (RHS)
D B
= A
HAHS Geometry Proofs ~ 2010
12
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.
EF=½BC (interval joining midpoints of sides of ABC is half the length 3rd side) Similarly in BFC , GH=½BC EF = GH
A
(Note: It can also be proven that EF and GH are parallel) E
F G
H C
B
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.
x 20 (interval parallel to side of ABC divides 9 15 other sides in same ratio) x = 12
A 15
9 J
>
I 20
x
>
B
C
Parallel lines preserve the ratio of intercepts on transversals. (converse is not true) Find the value of x.
> x
> 32
HAHS Geometry Proofs ~ 2010
>
x 18 (parallel lines preserve the ratios of 32 24 intercepts on transversals) * x = 24 18 24
* intercepts on parallel lines are in the same ratio * intercepts on parallel lines are in proportion
13
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.
x 2 122 152 (Pythagoras’ Theorem) x 2 225 144 81 x9
Find the value of x.
15 x
or
x 9 (3,4,5 Pythagorean Triad)
12 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) BC 2 10 2
Prove that ABC is right-angled
100
B
AB 2 AC 2 6 2 8 2 36 64 100
10 cm 6 cm
A
8 cm
HAHS Geometry Proofs ~ 2010
C
AB 2 AC 2 BC 2 ABC is right-angled (Converse of Pythagoras’ theorem)
14
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
15
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
16
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.
x = 68 (equal chords subtend equal angles at the centre)
A E 68
B
O
x
F Equal arcs subtend equal angles at the centre of a circle. (converse is true) arc AB = arc EF. Find the value of x.
x = 68 (equal arcs subtend equal angles at the centre)
A E 68
B
O
x
F Equal angles at the centre of a circle subtend equal chords. (converse is true) Chord EF = 16cm, find the length of chord AB.
A
AB = 16 cm (equal angles at the centre subtend equal chords)
E
75 B
HAHS Geometry Proofs ~ 2010
O
75 F
17
Equal angles at the centre of a circle subtend equal arcs. (converse is true) arc EF = 16cm, find the length of arc AB.
A
arc AB = 16 cm (equal angles at the centre subtend equal arcs)
E 75
O
75
B
16 cm F
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.
AP = 8 cm (interval through center perpendicular to chord AB bisects the chord)
B 8 cm P O
A 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.
interval through centre bisecting OEB 90 chord BC is perpendicular to the chord
O 6 cm
C
6 cm
B
E
NOTE: It can be proven that the perpendicular bisector of a chord passes through the centre of the circle.
HAHS Geometry Proofs ~ 2010
18
Chords equidistant from the centre of a circle are equal. (converse is true) Find the length of XY.
A
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)
P
5cm
B
= O =
Y
Q X Equal chords are equidistant from the centre of a circle. (converse is true) Find the length of OL.
IH = FG = 14 OL = 10 (equal chords are the equidistant from the centre)
I
F
7 7
M
L 7
5 7
O H
G
HAHS Geometry Proofs ~ 2010
19
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.
(i) y = 108 (angle at centre equals twice angle circumference standing on arc AB)
A
Note: use arc AB and not chord AB – the statement is not necessarily true for chords
54 O y C
B
(ii) Find the value of x.
(ii) x = 47 (angle at circumference equals half angle at centre standing on arc AB)
A x O 94 C
B
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.
x = 41 (angles at the circumference on the same arc PQ are equal) R
S
(Note: use arc PQ and not chord PQ – the statement is not necessarily true for chords)
x 41
or
O
x = 41 (angles at the circumference in the same segment equal) Q P
HAHS Geometry Proofs ~ 2010
20
Equal arcs subtend equal angles at the circumference. (converse is true) arc AB = arc CD. Find the value of x.
F
x = 37 (Equal arcs subtend equal angles at the circumference)
E Note: the statement is not necessarily true for equal chords
37
A
x
B D C Equal angles at the circumference subtend equal arcs. Find the length of arc PQ.
PQ = 8 cm (Equal angles at the circumference subtend equal arcs)
X 25
N
Y
8 cm 25
M
P
Q
The angle at the circumference in a semi-circle is 90o.
APˆ B 90 (angle at circumference in semi-circle equals 90o) x + 128 = 180 (angle sum of APB equals 180o) x = 52
AB is a diameter. Find the value of x.
P
x
38
A
O
B
A right angle at the circumference subtends a diameter If ACˆ B 90 then AB is a diameter. C
A
HAHS Geometry Proofs ~ 2010
B
21
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.
O 26
OTU = 90o (radius is perpendicular to tangent at point of contact) TOU + 116o = 180o (angle sum of OUT equals 180o) TOU = 64o
U
T S The angle between a tangent and a chord equals the angle at the circumference in the alternate segment. Find the size of RTN.
RTN = 93o (angle between tangent and chord equals angle at circumference in alternate segment)
R S
* Alternate segment theorem
93
M T N
HAHS Geometry Proofs ~ 2010
22
Cyclic Quadrilaterals The opposite angles of a cyclic quadrilateral are supplementary. (converse is true) Find the value of x.
x + 87 = 180 (opposite angles of cyclic quadrilateral ABCD are supplementary) x = 93
A 87
o
B * opposite angles of a cyclic quadrilateral are supplementary
D xo C The exterior angle of a cyclic quadrilateral equals the opposite (or remote) interior angle. (converse is true) Find the size of ADE.
ADE = 112o (exterior angle of cyclic quadrilateral ABCD equals opposite interior angle)
C
D
E
or ADE = 112o (exterior angle of cyclic quadrilateral ABCD equals remote interior angle)
o
112
B A
HAHS Geometry Proofs ~ 2010
23
Intercept Theorems The product of the intercepts on intersecting chords are equal. (converse is true) x 8 = 12 18 (product of intercepts on intersecting chords are equal) x = 27
Find the value of x. P B
12
8
x 18 A
Q
The product of the intercepts on intersecting secants are equal.
x 9 9 15 12 (product of intercepts on
Find the value of x.
intersecting secants are equal)
12
Q
P
T
3
9
9x + 81 = 180 9x = 99 x = 11
B x A
The square of the intercept on tangent to a circle equals the product of the intercepts on the secant.
x 2 16 4 (square of intercept on tangent to circle equals product of intercepts on secant) x2 = 64 x = 8 (length > 0)
Find the value of x. T x
B
4
P
12 A
HAHS Geometry Proofs ~ 2010
24
Intercepts on tangents drawn from a point to a circle are equal. Find the value of x.
x = 35 (intercepts on tangents from a point to a circle are equal) x
35
The line joining the centers of two circles passes through their point of contact
HAHS Geometry Proofs ~ 2010
25
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 YBZ = 90o (YB is an altitude) is a cyclic quadrilateral. XAZ = 90o (XA is an altitude) PBZ + PAZ = 180o Z AZBP is cyclic (opposite angles are B supplementary) X
P
A
Y
If the exterior angle of a quadrilateral equals the opposite interior angle then the quadrilateral is cyclic. Prove that ABCD is a cyclic quadrilateral. T
C
DAB = TCB (both 87o) ABCD is a cyclic (exterior angle equals opposite interior angle)
o
87
D
87
o
A
B
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 XBY = 90o (YB is an altitude) are the vertices of a cyclic quadrilateral. XAY = 90o (XA is an altitude) XBA = XAY = 90o Z XBAY is cyclic (XY subtends equal angles on B the same side at A and B) X
P
A
Y
HAHS Geometry Proofs ~ 2010
26
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 C
AF FB DF FC 36 A, C, B and D are concyclic (product of intercepts are equal)
6
9
F 6
4 B
D
HAHS Geometry Proofs ~ 2010
27
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