# Circle Theorems Euclid of Alexandria Circa 325 - 265 BC O The library of Alexandria was the foremost seat of learning in the world and functioned like.

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Circle Theorems Euclid of Alexandria Circa 325 - 265 BC O The library of Alexandria was the foremost seat of learning in the world and functioned like a university. The library contained 600 000 manuscripts.

diameter Circumference radius chord Major Segment Minor Segment Minor Arc Major Arc Minor Sector Major Sector A Reminder about parts of the Circle Tangent Parts

o Arc AB subtends angle x at the centre. A B xoxo Arc AB subtends angle y at the circumference. yoyo Chord AB also subtends angle x at the centre. Chord AB also subtends angle y at the circumference. o A B xoxo yoyo o yoyo xoxo A B Introductory Terminology Term’gy

Theorem 1 Measure the angles at the centre and circumference and make a conjecture. xoxo yoyo xoxo yoyo xoxo yoyo xoxo yoyo xoxo yoyo xoxo yoyo xoxo yoyo xoxo yoyo o o o o o o o o Th1

The angle subtended at the centre of a circle (by an arc or chord) is twice the angle subtended at the circumference by the same arc or chord. (angle at centre) 2x o Theorem 1 Measure the angles at the centre and circumference and make a conjecture. xoxo xoxo xoxo xoxo xoxo xoxo xoxo xoxo o o o o o o o o Angle x is subtended in the minor segment. Watch for this one later.

o A B 84 o xoxo Example Questions 1 Find the unknown angles giving reasons for your answers. o A B yoyo 2 35 o 42 o (Angle at the centre). 70 o (Angle at the centre) angle x = angle y =

(180 – 2 x 42) = 96 o (Isos triangle/angle sum triangle). 48 o (Angle at the centre) angle x = angle y = o A B 42 o xoxo Example Questions 3 Find the unknown angles giving reasons for your answers. o A B popo 4 62 o yoyo qoqo 124 o (Angle at the centre) (180 – 124)/2 = 28 0 (Isos triangle/angle sum triangle). angle p = angle q =

o Diameter 90 o angle in a semi-circle 20 o angle sum triangle 90 o angle in a semi-circle o a b c 70 o d 30 o e Find the unknown angles below stating a reason. angle a = angle b = angle c = angle d = angle e = 60 o angle sum triangle The angle in a semi-circle is a right angle.Theorem 2 This is just a special case of Theorem 1 and is referred to as a theorem for convenience. Th2

Angles subtended by an arc or chord in the same segment are equal. Theorem 3 xoxo xoxo xoxo xoxo xoxo yoyo yoyo Th3

38 o xoxo yoyo 30 o xoxo yoyo 40 o Angles subtended by an arc or chord in the same segment are equal. Theorem 3 Find the unknown angles in each case Angle x = angle y = 38 o Angle x = 30 o Angle y = 40 o

The angle between a tangent and a radius is 90 o. (Tan/rad) Theorem 4 o Th4

The angle between a tangent and a radius is 90 o. (Tan/rad) Theorem 4 o

180 – (90 + 36) = 54 o Tan/rad and angle sum of triangle. 90 o angle in a semi-circle 60 o angle sum triangle angle x = angle y = angle z = T o 36 o xoxo yoyo zozo 30 o A B If OT is a radius and AB is a tangent, find the unknown angles, giving reasons for your answers.

The Alternate Segment Theorem.Theorem 5 The angle between a tangent and a chord through the point of contact is equal to the angle subtended by that chord in the alternate segment. xoxo xoxo yoyo yoyo 45 o (Alt Seg) 60 o (Alt Seg) 75 o angle sum triangle 45 o xoxo yoyo 60 o zozo Find the missing angles below giving reasons in each case. angle x = angle y = angle z = Th5

Cyclic Quadrilateral Theorem.Theorem 6 The opposite angles of a cyclic quadrilateral are supplementary. (They sum to 180 o ) w x y z Angles x + w = 180 o Angles y + z = 180 o q p r s Angles p + q = 180 o Angles r + s = 180 o Th6

180 – 85 = 95 o (cyclic quad) 180 – 110 = 70 o (cyclic quad) Cyclic Quadrilateral Theorem.Theorem 6 The opposite angles of a cyclic quadrilateral are supplementary. (They sum to 180 o ) 85 o 110 o x y 70 o 135 o p r q Find the missing angles below given reasons in each case. angle x = angle y = angle p = angle q = angle r = 180 – 135 = 45 o (straight line) 180 – 70 = 110 o (cyclic quad) 180 – 45 = 135 o (cyclic quad)

Two Tangent Theorem.Theorem 7 From any point outside a circle only two tangents can be drawn and they are equal in length. P T U Q R PT = PQ P T U Q R Th7

90 o (tan/rad) Two Tangent Theorem.Theorem 7 From any point outside a circle only two tangents can be drawn and they are equal in length. P T Q O xoxo wowo 98 o yoyo zozo PQ and PT are tangents to a circle with centre O. Find the unknown angles giving reasons. angle w = angle x = angle y = angle z = 90 o (tan/rad) 49 o (angle at centre) 360 o – 278 = 82 o (quadrilateral)

90 o (tan/rad) Two Tangent Theorem.Theorem 7 From any point outside a circle only two tangents can be drawn and they are equal in length. P T Q O yoyo 50 o xoxo 80 o PQ and PT are tangents to a circle with centre O. Find the unknown angles giving reasons. angle w = angle x = angle y = angle z = 180 – 140 = 40 o (angles sum tri) 50 o (isos triangle) 50 o (alt seg) wowo zozo

O S T 3 cm 8 cm Find length OS OS = 5 cm (pythag triple: 3,4,5) Chord Bisector Theorem.Theorem 8 A line drawn perpendicular to a chord and passing through the centre of a circle, bisects the chord.. O Th8

Angle SOT = 22 o (symmetry/congruenncy) Find angle x O S T 22 o xoxo U Angle x = 180 – 112 = 68 o (angle sum triangle) Chord Bisector Theorem.Theorem 8 A line drawn perpendicular to a chord and passing through the centre of a circle, bisects the chord.. O

O S T 65 o P R U Mixed Questions PTR is a tangent line to the circle at T. Find angles SUT, SOT, OTS and OST. Angle SUT = Angle SOT = Angle OTS = Angle OST = 65 o (Alt seg) 130 o (angle at centre) 25 o (tan rad) 25 o (isos triangle) Mixed Q 1

22 o (cyclic quad) 68 o (tan rad) 44 o (isos triangle) 68 o (alt seg) Angle w = Angle x = Angle y = Angle z = O w y 48 o 110 o U Mixed Questions PR and PQ are tangents to the circle. Find the missing angles giving reasons. x z P Q R Mixed Q 2

He was 40 years old before he looked in on Geometry, which happened accidentally. Being in a Gentleman’s library, Euclid’s Elements lay open and twas the 47 El libri 1. He read the proposition. By God sayd he (he would now and then swear an emphaticall Oath by way of emphasis) this is impossible! So he reads the Demonstration of it which referred him back to such a Proposition, which proposition he read. That referred him back to another which he also read. Et sic deinceps that at last he was demonstratively convinced of the trueth. This made him in love with Geometry. From the life of Thomas Hobbes in John Aubrey’s Brief Lives, about 1694 Thomas Hobbes: Philosopher and scientist (1588 – 1679) Geometric Proofs

…"He studied and nearly mastered the Six-books of Euclid (geometry) since he was a member of Congress. He began a course of rigid mental discipline with the intent to improve his faculties, especially his powers of logic and language. Hence his fondness for Euclid, which he carried with him on the circuit till he could demonstrate with ease all the propositions in the six books; often studying far into the night, with a candle near his pillow, while his fellow-lawyers, half a dozen in a room, filled the air with interminable snoring.“…. (Abraham Lincoln from Short Autobiography of 1860.) Abraham Lincoln: 16 th U.S. President (1809 – 65)

At the age of twelve I experienced a second wonder of a totally different nature: in a little book dealing with Euclidean plane geometry, which came into my hands at the beginning of a school year. Here were assertions as for example, the intersection of the 3 altitudes of a triangle in one point, which– though by no means evident, could nevertheless be proved with such certainty that any doubt appeared to be out of the question. This lucidity and certainty, made an indescribable impression upon me. For example I remember that an uncle told me the Pythagorean Theorem before the holy geometry booklet had come into my hands. After much effort I succeeded in “proving” this theorem on the basis of similarity of triangles. For anyone who experiences [these feelings] for the first time, it is marvellous enough that man is capable at all to reach such a degree of certainty and purity in pure thinking as the Greeks showed us for the first time to be possible in geometry. From pp 9-11 in the opening autobiographical sketch of Albert Einstein: Philosopher – Scientist, edited by Paul Arthur.Schillp, published 1951 Albert Einstein E = mc 2

Extend AO to D AO = BO = CO (radii of same circle) Triangle AOB is isosceles (base angles equal) D   Triangle AOC is isosceles (base angles equal)   Angle AOB = 180 - 2  (angle sum triangle) Angle AOC = 180 - 2  (angle sum triangle) Angle COB = 360 – (AOB + AOC) (<‘s at point) Angle COB = 360 – (180 - 2  + 180 - 2  ) Angle COB = 2  + 2  = 2(  +  ) = 2 x < CAB To prove that angle COB = 2 x angle CAB QED To Prove that the angle subtended by an arc or chord at the centre of a circle is twice the angle subtended at the circumference by the same arc or chord. O C B A Theorem 1 and 2 Proof 1/2

O To Prove that angles subtended by an arc or chord in the same segment are equal. C B A Theorem 3 D To prove that angle CAB = angle BDC With centre of circle O draw lines OB and OC. Angle COB = 2 x angle CAB (Theorem 1). Angle COB = 2 x angle BDC (Theorem 1). 2 x angle CAB = 2 x angle BDC Angle CAB = angle BDC QED Proof 3

To prove that the angle between a tangent and a radius drawn to the point of contact is a right angle. Note that this proof is given primarily for your interest and completeness. Demonstration of the proof is beyond the GCSE course but is well worth looking at. The proofs up to now have been deductive proofs. That is they start with a premise, (a statement to be proven) followed by a chain of deductive reasoning that leads to the desired conclusion. The type of proof that follows is a little different and is known as “Reducto ad absurdum” It was first exploited with great success by ancient Greek mathematicians. The idea is to assume that the premise is not true and then apply a deductive argument that leads to an absurd or contradictory statement. The contradictory nature of the statement means that the “not true” premise is false and so the premise is proven true. Proof 4

To prove that OT is perpendicular to AB Assume that OT is not perpendicular to AB Then there must be a point, D say, on AB such that OD is perpendicular to AB. D C Since ODT is a right angle then angle OTD is acute (angle sum of a triangle). But the greater angle is opposite the greater side therefore OT is greater than OD. But OT = OC (radii of the same circle) therefore OC is also greater than OD, the part greater than the whole which is impossible. Therefore OD is not perpendicular to AB. By a similar argument neither is any other straight line except OT. Therefore OT is perpendicular to AB. QED To prove that the angle between a tangent and a radius drawn to the point of contact is a right angle. O A B T Theorem 4

To prove that the angle between a tangent and a chord through the point of contact is equal to the angle subtended by the chord in the alternate segment. Theorem 5 A T B C D O To prove that angle BTD = angle TCD With centre of circle O, draw straight lines OD and OT. Let angle DTB be denoted by .  Then angle DTO = 90 -  (Theorem 4 tan/rad) 90 -  Also angle TDO = 90 -  (Isos triangle) Therefore angle TOD = 180 –(90 -  + 90 -  ) = 2  (angle sum triangle) 22 Angle TCD =  (Theorem 1 angle at the centre)  Angle BTD = angle TCD QED Proof 5

To prove that angles A + C and B + D = 180 0 Draw straight lines AC and BD Chord DC subtends equal angles  (same segment)   Chord AD subtends equal angles  (same segment)   Chord AB subtends equal angles  (same segment)   Chord BC subtends equal angles  (same segment)   2(  +  +  +  ) = 360 o (Angle sum quadrilateral)  +  +  +  = 180 o Angles A + C and B + D = 180 0 QED A B D C To prove that the opposite angles of a cyclic quadrilateral are supplementary (Sum to 180 o ). Theorem 6     alphabeta gammadelta Proof 6

To prove that AP = BP. With centre of circle at O, draw straight lines OA and OB. To prove that the two tangents drawn from a point outside a circle are of equal length. Theorem 7 O A B P OA = OB (radii of the same circle) Angle PAO = PBO = 90 o (tangent radius). Draw straight line OP. In triangles OBP and OAP, OA = OB and OP is common to both. Triangles OBP and OAP are congruent (RHS) Therefore AP = BP. QED Proof 7

To prove that a line, drawn perpendicular to a chord and passing through the centre of a circle, bisects the chord. Theorem 8 O A B C To prove that AB = BC. From centre O draw straight lines OA and OC. In triangles OAB and OCB, OC = OA (radii of same circle) and OB is common to both. Angle OBA = angle OBC (angles on straight line) Triangles OAB and OCB are congruent (RHS) Therefore AB = BC QED Proof 8

Worksheet 1 Parts of the Circle

Worksheet 2 xoxo yoyo o xoxo yoyo o xoxo yoyo o xoxo yoyo o xoxo yoyo o xoxo yoyo o Th1 Measure the angle subtended at the centre (y) and the angle subtended at the circumference (x) in each case and make a conjecture about their relationship.

To Prove that the angle subtended by an arc or chord at the centre of a circle is twice the angle subtended at the circumference by the same arc or chord. O C B A Theorem 1 and 2 Worksheet 3

To Prove that angles subtended by an arc or chord in the same segment are equal. A Theorem 3 O C B D Worksheet 4

To prove that the angle between a tangent and a radius drawn to the point of contact is a right angle. O A B T Theorem 4 Worksheet 5

To prove that the angle between a tangent and a chord through the point of contact is equal to the angle subtended by the chord in the alternate segment. Theorem 5 A T B C D O Worksheet 6

A B D C To prove that the opposite angles of a cyclic quadrilateral are supplementary (Sum to 180 o ). Theorem 6     AlphaBeta Chidelta Worksheet 7

Worksheet 8 To prove that the two tangents drawn from a point outside a circle are of equal length. Theorem 7 O A B P

To prove that a line, drawn perpendicular to a chord and passing through the centre of a circle, bisects the chord. Theorem 8 O A B C Worksheet 9

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