OpenStax-CNX module: m39327 1 Geometry: Introduction, Circle Geometry (Grade 12) Free High School Science Texts Project This work is produced by OpenStax-CNX and licensed under the Creative Commons Attribution License 3.0 1 Introduction 1.1 Discussion : Discuss these Research Topics Research one of the following geometrical ideas and describe it to your group: 1. taxicab geometry, 2. spherical geometry, 3. fractals, 4. the Koch snowake. 2 Circle Geometry 2.1 Terminology The following is a recap of terms that are regularly used when referring to circles. arc: An arc is a part of the circumference of a circle. chord: A chord is dened as a straight line joining the ends of an arc. radius: The radius, r, is the distance from the centre of the circle to any point on the circumference. diameter: The diameter is a special chord that passes through the centre of the circle. The diameter is the straight line from a point on the circumference to another point on the circumference, that passes through the centre of the circle. segment: A segment is the part of the circle that is cut o by a chord. A chord divides a circle into two segments. tangent: A tangent is a line that makes contact with a circle at one point on the circumference. (AB is a tangent to the circle at point P ). Version 1.1: Aug 1, 2011 3:11 am -0500 http://creativecommons.org/licenses/by/3.0/
OpenStax-CNX module: m39327 2 Figure 1: Parts of a Circle 2.2 Axioms An axiom is an established or accepted principle. For this section, the following are accepted as axioms. 1. The Theorem of Pythagoras, which states that the square on the hypotenuse of a right-angled triangle is equal to the sum of the squares on the other two sides. In [U+25B5]ABC, this means that (AB) 2 + (BC) 2 (AC) 2 Figure 2: A right-angled triangle 2. A tangent is perpendicular to the radius, drawn at the point of contact with the circle. 2.3 Theorems of the Geometry of Circles A theorem is a general proposition that is not self-evident but is proved by reasoning (these proofs need not be learned for examination purposes). Theorem 1 The line drawn from the centre of a circle, perpendicular to a chord, bisects the chord. Figure 3 Consider a circle, with centre O. Draw a chord AB and draw a perpendicular line from the centre of the circle to intersect the chord at point P. The aim is to prove that AP BP 1. [U+25B5]OAP and [U+25B5]OBP are right-angled triangles. 2. OA OB as both of these are radii and OP is common to both triangles. Apply the Theorem of Pythagoras to each triangle, to get: OA 2 OP 2 + AP 2 OB 2 OP 2 + BP 2 (1)
OpenStax-CNX module: m39327 3 However, OA OB. So, OP 2 + AP 2 OP 2 + BP 2 AP 2 BP 2 and AP BP (2) This means that OP bisects AB. Theorem 2 The line drawn from the centre of a circle, that bisects a chord, is perpendicular to the chord. Figure 4 Consider a circle, with centre O. Draw a chord AB and draw a line from the centre of the circle to bisect the chord at point P. The aim is to prove that OP AB In [U+25B5]OAP and [U+25B5]OBP, 1. AP P B (given) 2. OA OB (radii) 3. OP is common to both triangles. [U+25B5]OAP [U+25B5]OBP (SSS). OP A OP B OP A + OP B 180 (APB is a str. line) OP A OP B 90 OP AB (3) Theorem 3 The perpendicular bisector of a chord passes through the centre of the circle. Figure 5 Consider a circle. Draw a chord AB. Draw a line P Q perpendicular to AB such that P Q bisects AB at point P. Draw lines AQ and BQ. The aim is to prove that Q is the centre of the circle, by showing that AQ BQ. In [U+25B5]OAP and [U+25B5]OBP, 1. AP P B (given) 2. QP A QP B (QP AB) 3. QP is common to both triangles.
OpenStax-CNX module: m39327 4 [U+25B5]QAP [U+25B5]QBP (SAS). From this, QA QB. Since the centre of a circle is the only point inside a circle that has points on the circumference at an equal distance from it, Q must be the centre of the circle. 2.3.1 Circles I 1. Find the value of x: Figure 6 Figure 7 Theorem 4 The angle subtended by an arc at the centre of a circle is double the size of the angle subtended by the same arc at the circumference of the circle. Figure 8 Consider a circle, with centre O and with A and B on the circumference. Draw a chord AB. Draw radii OA and OB. Select any point P on the circumference of the circle. Draw lines P A and P B. Draw P O and extend to R. The aim is to prove that AOB 2 AP B. AOR P AO + AP O (exterior angle sum of interior opp. angles) But, P AO AP O ([U+25B5]AOP is an isosceles [U+25B5]) AOR 2 AP O Similarly, BOR 2 BP O. So, AOB AOR + BOR 2 AP O +2 BP O ( ) 2 AP O + BP O ) ( 2 AP B (4)
OpenStax-CNX module: m39327 5 2.3.2 Circles II 1. Find the angles (a to f ) indicated in each diagram: Figure 9 Theorem 5 The angles subtended by a chord at the circumference of a circle on the same side of the chord are equal. Figure 10 Consider a circle, with centre O. Draw a chord AB. Select any points P and Q on the circumference of the circle, such that both P and Q are on the same side of the chord. Draw lines P A, P B, QA and QB. The aim is to prove that AQB AP B. and 2 AOB 2 AQB at centre twice at circumference AOB 2 AP B at centre twice at circumference AQB 2 AP B AQB AP B (5) Theorem 6 (Converse of Theorem ) If a line segment subtends equal angles at two other points on the same side of the line, then these four points lie on a circle. Figure 11 Consider a line segment AB, that subtends equal angles at points P and Q on the same side of AB. The aim is to prove that points A, B, P and Q lie on the circumference of a circle. By contradiction. Assume that point P does not lie on a circle drawn through points A, B and Q. Let the circle cut AP (or
OpenStax-CNX module: m39327 6 AP extended) at point R. but but this cannot be true since AQB AQB ARB ARB ARB 's on same side of chord AP B + RBP AP B (given) AP B (ext. of [U+25B5]) (6) the assumption that the circle does not pass through P, must be false, and A, B, P and Q lie on the circumference of a circle. 2.3.3 Circles III 1. Find the values of the unknown letters. Figure 12 2.3.4 Cyclic Quadrilaterals Cyclic quadrilaterals are quadrilaterals with all four vertices lying on the circumference of a circle. vertices of a cyclic quadrilateral are said to be concyclic. Theorem 7 The opposite angles of a cyclic quadrilateral are supplementary. The Figure 13 Consider a circle, with centre O. Draw a cyclic quadrilateral ABP Q. Draw AO and P O. The aim is
OpenStax-CNX module: m39327 7 to prove that ABP + AQP 180 and QAB + QP B 180. 2 But, O 1 2 ABP 's at centre O 2 2 AQP 's at centre O 1 + O 2 360 ABP +2 AQP 360 ABP + AQP 180 Similarly, QAB + QPB 180 (7) Theorem 8 (Converse of Theorem ) If the opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic. Figure 14 Consider a quadrilateral ABP Q, such that ABP + AQP 180 and QAB + QP B 180. The aim is to prove that points A, B, P and Q lie on the circumference of a circle. By contradiction. Assume that point P does not lie on a circle drawn through points A, B and Q. Let the circle cut AP (or AP extended) at point R. Draw BR. but but this cannot be true since QAB + QRB 180 opp. 's of cyclic quad QAB + QPB 180 (given) QRB QRB QP B QP B + RBP (ext. of[u+25b5]) (8) the assumption that the circle does not pass through P, must be false, and A, B, P and Q lie on the circumference of a circle and ABP Q is a cyclic quadrilateral. 2.3.4.1 Circles IV 1. Find the values of the unknown letters. Figure 15
OpenStax-CNX module: m39327 8 Theorem 9 Two tangents drawn to a circle from the same point outside the circle are equal in length. Figure 16 Consider a circle, with centre O. Choose a point P outside the circle. Draw two tangents to the circle from point P, that meet the circle at A and B. Draw lines OA, OB and OP. The aim is to prove that AP BP. In [U+25B5]OAP and [U+25B5]OBP, 1. OA OB (radii) 2. OAP OP B 90 (OA AP and OB BP ) 3. OP is common to both triangles. [U+25B5]OAP [U+25B5]OBP (right angle, hypotenuse, side) AP BP 2.3.4.2 Circles V 1. Find the value of the unknown lengths. Figure 17 Theorem 10 The angle between a tangent and a chord, drawn at the point of contact of the chord, is equal to the angle which the chord subtends in the alternate segment. Figure 18 Consider a circle, with centre O. Draw a chord AB and a tangent SR to the circle at point B. Chord AB subtends angles at points P and Q on the minor and major arcs, respectively. Draw a diameter BT and join A to T. The aim is to prove that AP B ABR and AQB ABS. First prove that AQB ABS as
OpenStax-CNX module: m39327 9 this result is needed to prove that AP B ABR. ABS + ABT 90 (TB SR) BAT 90 ( 's at centre) ABT + AT B 90 (sum of angles in [U+25B5]BAT) However, ABS AQB AQB AT B ABT (angles subtended by same chord AB) ABS (9) SBQ + QBR 180 (SBT is a str. line) AP B + AQB 180 (ABPQ is a cyclic quad) SBQ + QBR AQB AP B AP B + AQB ABS ABR 2.3.4.3 Circles VI 1. Find the values of the unknown letters. Figure 19 Theorem 11 (Converse of ) If the angle formed between a line, that is drawn through the end point of a chord, and the chord, is equal to the angle subtended by the chord in the alternate segment, then the line is a tangent to the circle. Figure 20 Consider a circle, with centre O and chord AB. Let line SR pass through point B. Chord AB subtends an angle at point Q such that ABS AQB. The aim is to prove that SBR is a tangent to the circle.
OpenStax-CNX module: m39327 10 By contradiction. Assume that SBR is not a tangent to the circle and draw XBY such that XBY is a tangent to the circle. However, But, can only be true if, ABX ABS ABX ABX AQB (tan chord theorem) AQB ABS (given) ABS + XBS XBS 0 (10) If XBS is zero, then both XBY and SBR coincide and SBR is a tangent to the circle. 2.3.4.4 Applying Theorem 1. Show that Theorem also applies to the following two cases: Figure 21 Exercise 1: Circle Geometry I (Solution on p. 12.) Figure 22 BD is a tangent to the circle with centre O. BO AD. Prove that: 1. CF OE is a cyclic quadrilateral 2. F B BC 3. [U+25B5]COE///[U+25B5]CBF 4. CD 2 ED.AD 5. OE BC CD CO Exercise 2: Circle Geometry II (Solution on p. 13.)
OpenStax-CNX module: m39327 11 Figure 23 F D is drawn parallel to the tangent CB Prove that: 1. F ADE is cyclic 2. [U+25B5]AF E [U+25B5]CBD 3. F C.AG GH DC.F E BD
OpenStax-CNX module: m39327 12 Solutions to Exercises in this Module Solution to Exercise (p. 10) Step 1. Step 2. Let OEC x. F OE 90 (BO OD) F CE 90 ( subtended by diameter AE) CF OE is a cyclic quad (opposite 's supplementary) (11) F CB x ( between tangentbd and chord CE) BF C x (exterior to cyclic quad CFOE) BF BC (sides opposite equal 's in isosceles [U+25B5]BFC) (12) Step 3. CBF 180 2x (sum of 's in [U+25B5]BFC) OC OE (radii of circle O) ECO x (isosceles [U+25B5]COE) COE 180 2x (sum of 's in [U+25B5]COE) (13) COE CBF ECO F CB OEC CF B [U+25B5]COE [U+25B5]CBF (3 's equal) (14) Step Step 4. a. In [U+25B5]EDC Step b. In [U+25B5]ADC CED 180 x ( 's on a str. linead) ECD 90 x (complementary 's) ( ACE 180 x sum of 's ACE and ) ECO CAD 90 x (sum of 's in [U+25B5]CAE) (15) (16) Step c. Lastly, ADC EDC since they are the same.
OpenStax-CNX module: m39327 13 Step d. [U+25B5]ADC [U+25B5]CDE (3 's equal) ED CD CD AD CD 2 ED.AD (17) Step Step 5. a. Step b. In [U+25B5]BCO OE CD ([U+25B5]OEC is isosceles) (18) Step c. In [U+25B5]OCD OCB 90 (radius OC on tangent BD) CBO 180 2x (sum of 's in [U+25B5]BFC) OCD 90 (radius OC on tangent BD) COD 180 2x (sum of 's in [U+25B5]OCE) (19) (20) Step d. Lastly, OC is a common side to both [U+25B5]'s. Step e. [U+25B5]BOC [U+25B5]ODC (common side and 2 equal angles) OE BC CD CO CO BC CD CO (OE CD isosceles [U+25B5]OEC) (21) Solution to Exercise (p. 10) Step 1. Let BCD x Step Step 2. a. Let F EA y CAH x ( between tangent BC and chord CE) F DC x (alternate, FD CB) FADE is a cyclic quad (chord FE subtends equal 's) (22) F DA y ( 's subtended by same chordaf in cyclic quad FADE) CBD y (corresponding 's, FD CB) F EA CBD (23) Step b. BCD F AE (above) (24) Step c. AF E 180 x y ( 's in [U+25B5]AFE) CBD 180 x y ( 's in [U+25B5]CBD) [U+25B5]AF E [U+25B5]CBD (3 's equal) (25)
OpenStax-CNX module: m39327 14 Step Step 3. a. Step b. Step c. AG GH F A F C DC BD F A F E DC.F E BD F A (26) (FG CH splits up lines AH and AC proportionally) F C.AG GH F A F C.AG GH DC.F E BD (27) (28)