Geometry theorems

Question 1

what is the radius of the circle

Answer 1

a line from the Centre to any point on the circumference of the circle

Question 2

what is the diameter of the circle

Answer 2

a line passing through the center of the circle. It is double the length of the radius.

Question 3

what is the chord of the circle

Answer 3

a line with end-points on the circumference .

Question 4

what is the secant of the circle

Answer 4

a line passing through 2 points on the circle

Question 5

draw a circle with the different labeling

Answer 5

see workbook

Question 6

what is a tangent of the circle

Answer 6

line that touches the circle at exactly one point, never entering the circle’s interior.

Question 7

what is a sector of the circle

Answer 7

is the portion of a disk enclosed by two radii and an arc.

Question 8

what is a segment of the circle

Answer 8

region bounded by a chord and a corresponding arc lying between the chord’s endpoints. In other words, a circular segment is a region of a circle which is created by breaking apart from the rest of the circle through a secant or a chord.

Question 9

what is a arc of the circle

Answer 9

a portion of the circumference of a circle. Or. A piece of a circle between two points is also called an arc.

Question 10

State theorem 1

Answer 10

a line drawn perpendicular from the center of a circle to chord bisects the cord.

Question 11

Reasoning for theorem 1

Answer 11

line from centre ⊥ to chord

Question 12

How do you prove theorem 1

Answer 12

see workbook

Given: Circle with center O with OM ⊥ AB.

What to prove: AM = MB

Construction: Join OA and OB

Proof: In ∆ OAM and ∆ OBM:

(i) OA = OB radii
(ii) 𝑀1 ̂ = 𝑀 ̂2 = 90° given
(iii) OM = OM common
∴ ∆ OAM ≡ ∆ OBM (RHS)

∴ AM = MB

Question 13

state the converse of theorem 1

Answer 13

line from center to midpoint of chord

Question 14

prove converse of theorem 1

Answer 14

see workbook

Given: Circle with centre O. M is a point on chord AB such that AM = MB.

What to prove: OM ⊥ AB

Construction: Join OA and OB

Proof: In ∆ OAM and ∆ OBM:

(i) OA = OB radii
(ii) AM = BM given
(iii) OM = OM common
∴ ∆ OAM ≡ ∆ OBM (SSS)

∴ 𝑀1 ̂ = 𝑀 ̂2 = 90° ∠𝑠 on straight line

Question 15

State theorem 2

Answer 15

The angle with an arc of a circle subtends at the center of a circle is twice the angle it subtends at the circumference of the circle.

Question 16

draw 2 other variations of theorem 2

Answer 16

see workbook

Question 17

reasoning for theorem 2

Answer 17

angle at center is 2x the angle at the circumference.

Question 18

proof for theorem 2

Answer 18

see workbook
Proof: (i)
Let C1 ̂ = 𝑥 and 𝐶2 ̂ = 𝑦
C1 ̂= A ̂ = 𝑥 ∠s opp. equal radii

O1 ̂ = C1 ̂+ A ̂ = 2𝑥 Ext ∠ of ΔOAC

Similarly, in 𝛥 OCB: O2 ̂ = C2 ̂+ B ̂ = 2𝑦

O1 ̂ + O2 ̂ = 2𝑥 + 2𝑦 =𝟐(𝒙 + 𝒚)
= 𝟐 (C1 ̂+ C2 ̂)
∴ AO ̂B = 2 × 𝐴C ̂B

Question 19

what is a semi-circle

Answer 19

half of a circle ;the arc from one end of a diameter to the other

Question 20

what does it mean to be subtended

Answer 20

in geometry, an angel is subtended by an arc, line segment or any other section of a curve when its 2 rays pass through the endpoints of that arc, line segment or curve section

Question 21

State theorem 3

Answer 21

The angle subtended at the circle by a diameter is a right angle (angle in a semi circle is 90 degrees)

Question 22

reasoning for theorem 3

Answer 22

angles in a semi circle

Question 23

State theorem 4

Answer 23

an arc or chord of a circle subtends equal angles at the circumference of the circle .(angles in the same segment of the circle are equal is subtended by the same arc/chord).

Question 24

reasoning for theorem 4

Answer 24

angles in the same segment

Question 25

state theorem 5

Answer 25

equal chords subtend equal angles at the circumference.

Question 26

reasoning for theorem 5

Answer 26

equal chord=equal angles

Question 27

State theorem 6

Answer 27

the opposite angles of a cyclic quadrilateral are supplementary.

Question 28

reasoning for theorem 6

Answer 28

opposite angles of a cyclic quad

Question 29

State the converse of theorem 6

Answer 29

if the opposite angles of a quadrilateral are supplementary then the quadrilateral is a cyclic quadrilateral

Question 30

reasoning for theorem 6 converse

Answer 30

opposite angles of cyclic quad

Question 31

proof for theorem 6

Answer 31

see workbook Proof: Let D ̂ = 𝑥 and B ̂ = y 𝑂 ̂1 = 2𝑥 ∠ at centre = 2× ∠ at circumference 𝑂 ̂2= 2𝑦 ∠ at centre = 2× ∠ at circumference 𝑂 ̂1 + 𝑂 ̂2 = 360° revolution 2𝑥 + 2𝑦 = 360° 2(𝑥 + 𝑦) = 360° 𝑥 + 𝑦 = 180° ∴ 𝐷 ̂ + 𝐵 ̂ = 180° Similarly, by joining BO and DO, it can be proven that 𝐴 ̂ + 𝐶 ̂ = 180°

Question 32

State theorem 7

Answer 32

an exterior angle of a cyclic quad is equal to the interior opposite angle

Question 33

State the converse of Theorem 7

Answer 33

if an exterior angle of a quadrilateral is equal to the interior opposite angle, then the quadrilateral is a quadrilateral

Question 34

reasoning for theorem 7

Answer 34

exterior angle f cyclic quad

Question 35

State theorem 8

Answer 35

A tangent to a circle is perpendicular to the radius at the point of contact.

Question 36

State the converse of Theorem 8

Answer 36

If a line is drawn perpendicular to a radius at the point where the radius meets the circle, then the line is a tangent to the circle

Question 37

reasoning for theorem 8

Answer 37

tan ⊥ radius

Question 38

reasoning for converse of theorem 8

Answer 38

tan ⊥ radius

Question 39

State theorem 9

Answer 39

The angle between a tangent to a circle and a chord drawn from the point of contact is equal to an angle in the alternate segment

Question 40

State the converse of theorem 9

Answer 40

If a line is drawn through the endpoint of a chord, making with the chord an angle equal to an angle in the alternate segment, then the line is a tangent to the circle.

Question 41

State the reasoning for theorem 9

Answer 41

Tan-chord theorem

Question 42

State the reasoning for converse of theorem 9

Answer 42

tan-chord theorem

Question 43

proof for theorem 9

Answer 43

Proof: 𝐵 ̂1 + 𝐵 ̂2 = 90° tan ⊥ radius 𝐸 ̂1 + 𝐸 ̂2 = 90° ∠ in semi-circle Let B1 ̂ = 𝑥 ∴ 𝐵 ̂2 = 90° − 𝑥 𝐵 ̂1= 𝐸 ̂1 = 𝑥 ∠ s in the same segment ∴ 𝐸 ̂2 = 90° − 𝑥 ∴ 𝐵2 ̂ = 𝐸2 ̂

Question 44

State theorem 10

Answer 44

If two tangents are drawn from the same point outside a circle, then they are equal in length.

Question 45

reasoning for theorem 10

Answer 45

Tans form the same point A