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H3 Mathematics Plane Geometry 1
Chapter 4
Circles, Tangent-Chord Theorem, Intersecting Chord Theorem and Tangent-secant
Theorem
Outline
• Basic definitions and facts on circles
• The Tangent-Chord Theorem
• The Intersecting Chord Theorem
• The Tangent – Secant Theorem
4.1 Basic definitions and facts on circles
Example 1 Given the circle below
The center is O, AO is a radius, AM is a diameter, NK is a chord and
AM is the arc intercepted by the central angle AOM∠
Measurement of angles
Consider a protractor used to measure angles. One divides the circle into 360 equal parts. To measure the
central angle, one measures actually the arc intercepted by the central angle. Each unit of arc is 1 arc degree
which has 1o
central angle.
Postulates
(a) A central angle has the same number of degree as its intercepted arc.
(b) In the same or equal circle(s) equal central angles intercept equal arc.
(c) In the same or equal circle(s) equal arcs are intercepted by equal central angles.
(d) In the same or equal circle(s), equal chords are equally distant from the center.
(e) In the same or equal circle(s), chords equally distant from the center are equal.
Theorem 1 An inscribed angle has half as many degree as the intercepted arc.
For example, if the arch BC is ao
, then BAC∠ is 2
1
ao
.
Tampines Junior College
H3 Mathematics (9810)
Plane Geometry
O
A
L M
N
K
O
B
A
C
A corollary is a mathematical statement which follows easily from a previously proven
statement.
A postulate is a proposition that requires no proof, being self-evident, or that is for a specific
purpose assumed true, and that is used in the proof of other propositions.
H3 Mathematics Plane Geometry 2
Corollary 1 An angle inscribed in a semicircle is a right angle.
4.2 Tangent of a circle
Definition 1
A tangent of a circle is a line that intersects the circle at exactly one point.
Theorem 2 A straight line perpendicular to a radius at its outer extremity is a tangent to the circle.
Given that OC is a radius and ACB is perpendicular to OC.
To prove AB is tangent to circle O.
The plan is to show that every point on AB different from C is not on the circle.
Proof
1. Let D be any point on AB different from C.
Connect OD
2. OC is perpendicular to AB (Given)
3. OD>OC (OD is the hypotenuse of the right triangle)
4. D is not on the circle.
5. AB is tangent to the circle (C is the only point on both AB and the circle).
Theorem 3 A tangent to a circle is perpendicular to the radius drawn to the point of contact.
Given in the figure that ACB is tangent to the circle O at point C.
To prove that OC ⊥ AB
Proof (We prove by contradiction)
1. Suppose OC is not perpendicular to AB
2. Choose a point D on AB such that OD ⊥ AB
(From any point a perpendicular line to a given line can be drawn)
3. Choose point E on AB such that CD = DE and E is on different side of D.
O
C
A
O
C
A
B
D
O
C
A
H3 Mathematics Plane Geometry 3
4. ∠ODC = ∠ODE = 90°
5. OD = OD (Same side)
6. ∆ ODC ≅ ∆ ODE (S.A.S.)
7. OC = OE (Corresponding sides of congruent triangles are equal)
8. E is on O. (by 7)
9. AB intersects the circle at two different points
10. So AB is not tangential to the circle.
The contradiction shows that OC ⊥ AB
In-Class Exercise 1
Suppose in the given figure that O is the center of the circle, AB and DC are tangents to the circle at B and C
respectively. Show that AC = DB.
Proof
1. ∠BOA = ∠COD (Vertically opposite angle)
2. OB = OC (radius of the circle)
3. ∠OBA = ∠OCD = 90° (Tangent is perpendicular to radius)
4. ∆ABO ≅ ∆DCO (A.S.A.)
5. AO = DO (Corresponding sides of congruent triangles are equal)
6. AO + OC = DO + OB (OB = OC = radius of the same circle)
7. AC = DB
Theorem 4 (Tangent-Chord Theorem)
The angle between a tangent and a chord meeting the tangent at the point of contact is equal to the inscribed
angle on opposite side of the chord.
Given that AB is tangent to the circle at C, CD is a chord, E is on the circle.
To prove ∠ DEC =∠ DCB
O
C
A
B
D
E
C
O
B A
ED
O
A
D
B C
H3 Mathematics Plane Geometry 4
Proof
1. Draw the chord FD // AB.
2. Let G be the intersection of ray CO and FD.
3. FD is perpendicular to CG (AB is perpendicular to CG)
4. ∠ DFC =∠ DEC (Intercept the same arc)
5. OF = OD (Radius of the same circle)
6. ∠ OFG =∠ ODG ( ∆ OFD is isosceles)
7. ∠ FGO = ∠ DGO = 90° (by 3)
8. ∆ OFG ≅ ∆ ODG (by 5, 6, 7 and A.S.A)
9. FG = GD (by 8)
10. ∠ FGC =∠ DGC = 90o
(by 3)
11. GC = GC
12. ∆ CFG ≅ ∆ CDG (by 9, 10, 11 and S.A.S.)
13. ∠ DFC =∠ FDC (corresponding angles of congruent triangles)
14. ∠ FDC =∠ DCB (alternative interior angles of parallel lines)
15. ∠ DCB =∠ DEC (by 4, 13, 14)
In-Class-Exercise 2
Given in the figure that ACB is a tangent of the circle, ∠ ECD = 75o
and ∠ DCB = 45o
. Find ∠ EDC.
Solution
∠EDC = 180° - ∠ DEC - ∠ ECD
= 180° - 45° - 75° (Tangent Chord Theorem)
= 60°
Remark The Tangent-chord theorem can be deduced informally from the fact that inscribed angles
that intercept the same arc equal. Imagine the point E in the
above figure moves continuously to C along the circle. Then
DE approaches CD and EC tends to be parallel to AB.
Thus the measurement of ∠ DEC approaches
to the measurement of ∠ DCB.
F
O
C
A
B
D
E
G
C
A
B
D
E
C
B
A
E
D
H3 Mathematics Plane Geometry 5
Example 1 Suppose in the figure line AB is tangential to the circle at C, DE //AB.
Prove (EC)(CG) = (CD)(CF).
Proof
1. ∠GFC = ∠GCB (Tangent Chord Theorem)
2. ∠ECB = ∠DEC (Alternate angles)
3. ∠GFC = ∠DEC (by 1 & 2)
4. ∠DCE = ∠GCF (Same angle)
5. ∆DCE ~ ∆GFC (A.A.)
6.
CG
CD
FC
EC
= (by 5)
7. (EC)(CG) = (CD)(CF) (by 6)
Remark
From example 1 we can create the following new problems. Can you see why?
(1) Suppose in the figure line AB tangents the circle at C, FE //AB.
Prove (EC)(CG) = (CF) 2
.
D coincides with F, hence CD = CF
(2) Suppose in the figure GK is a diameter, FJ is perpendicular to GK.
Prove (FG) 2
= (GJ)(GH).
FJ is perpendicular to the GK and GK is perpendicular to the
Tangent at G, hence FJ is parallel to the tangent at G.
Can you create any more new questions?
E
C
BA
F
G
D
E
C
A B
F
G
J
G
F
H
K
H3 Mathematics Plane Geometry 6
4.3 Intersecting Chord Theorem, Tangent-secant theorem
Theorem 5 (Intersecting Chord Theorem)
If two chords of a circle intersect in the interior of a circle, thus determine two segments in each chord, the
product of the lengths of the segments of one chord equals the product of the lengths of the segments of the
other chord.
Given that AB and CD are two chords intersecting at P.
To prove that (AP)(BP) = (CP)(DP).
The plan is to use properties of similar triangles.
Proof
1. Connect A and D, B and C.
2. ∠ ADP =∠ CBP (They intercept the same arc)
3. ∠ DAP =∠ BCP (As in 2)
4. ∆ ADP ~∆ CPB (A.A.)
5. AP: CP = DP: BP (by 4)
6. (AP)(BP) = (DP)(CP) (by 5)
In-Class-Activity
If two intersecting chords bisect each other, what can you say about their relationship?
Answer: They are the diameters of the circle
Definition 2
A secant of a circle is a line drawn from a point outside the circle that intersects the circle at two points.
For example, the line AB is a secant of the circle.
A secant is not a tangent.
Theorem 6 If from a point outside a circle two secants are drawn, the product of one secant and its external
segment is equal to the product of the other secant and its external segment.
Given that CBA and CDE are secants from the point C outside the circle.
To prove that (CB)(CA) = (CD)(CE).
The plan is to show∆ ADC ~∆ EBC
P
C
D
A
B
A
B
D
B C
A
E
P
A
B
D
C
H3 Mathematics Plane Geometry 7
Proof
1. ∠CEB = ∠DAB (They intercept the same arc)
2. ∠DCA = ∠DCA (Same angle)
3. ∆ECB ~ ∆ACD (A.A.)
4.
EC
AC
BC
CD
= (Ratio of corresponding sides of similar triangles)
5. (CB)(CA) = (CD)(CE) (by 4)
Theorem 7 (Tangent –Secant Theorem)
If from a point outside a circle a secant and a tangent are drawn, the secant and its external segment is equal
to the square of the tangent.
Given tangent AB and secant ACD are from an external point A.
To prove that (AB)2
= (AC)(AD).
The plan is to show that ∆ ACB ~ ∆ ABD.
Proof
1. ∠ CAB =∠ BAD (same angle)
2. ∠ CBA =∠ BDA (By Tangent-chord Theorem)
3. ∆ ACB ~ ∆ ABD (A.A)
4. AB: AD = AC: AB (by 3)
5. (AB)2
= (AC)(AD
Example 2 Suppose in the figure the arc DM equals arc ME and BC tangents the circle at M. Prove DB:
BM = EC: CM.
Proof
1. AM is the bisector of ∠BAC (Given)
2. BA: CA = BM: CM (Tutorial 3 qn 6)
3. BA: BM = CA: CM (by 2)
4. BM: BD = BA: BM and CM: CE = CA: CM (Tangent-Secant Theorem)
5. BM: BD = CM: CE (by 3 & 4)
6. DB: BM = EC: CM (by 5)
C
B
A
D
C
A
D
B
ED
M
A
CB
B
C
A
E
D
H3 Mathematics Plane Geometry 8
Summary
• Main terms
Inscribed angle, chord, radius, diameter, tangent, secant
• Main results
Tangent-Chord Theorem
Intersecting chord Theorem
Tangent-secant Theorem
• Useful facts
. An inscribed angle has half as many degrees as the intercepted arc.
. An angle inscribed in a semicircle is a right angle.

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Circlestangentchordtheorem

  • 1. H3 Mathematics Plane Geometry 1 Chapter 4 Circles, Tangent-Chord Theorem, Intersecting Chord Theorem and Tangent-secant Theorem Outline • Basic definitions and facts on circles • The Tangent-Chord Theorem • The Intersecting Chord Theorem • The Tangent – Secant Theorem 4.1 Basic definitions and facts on circles Example 1 Given the circle below The center is O, AO is a radius, AM is a diameter, NK is a chord and AM is the arc intercepted by the central angle AOM∠ Measurement of angles Consider a protractor used to measure angles. One divides the circle into 360 equal parts. To measure the central angle, one measures actually the arc intercepted by the central angle. Each unit of arc is 1 arc degree which has 1o central angle. Postulates (a) A central angle has the same number of degree as its intercepted arc. (b) In the same or equal circle(s) equal central angles intercept equal arc. (c) In the same or equal circle(s) equal arcs are intercepted by equal central angles. (d) In the same or equal circle(s), equal chords are equally distant from the center. (e) In the same or equal circle(s), chords equally distant from the center are equal. Theorem 1 An inscribed angle has half as many degree as the intercepted arc. For example, if the arch BC is ao , then BAC∠ is 2 1 ao . Tampines Junior College H3 Mathematics (9810) Plane Geometry O A L M N K O B A C A corollary is a mathematical statement which follows easily from a previously proven statement. A postulate is a proposition that requires no proof, being self-evident, or that is for a specific purpose assumed true, and that is used in the proof of other propositions.
  • 2. H3 Mathematics Plane Geometry 2 Corollary 1 An angle inscribed in a semicircle is a right angle. 4.2 Tangent of a circle Definition 1 A tangent of a circle is a line that intersects the circle at exactly one point. Theorem 2 A straight line perpendicular to a radius at its outer extremity is a tangent to the circle. Given that OC is a radius and ACB is perpendicular to OC. To prove AB is tangent to circle O. The plan is to show that every point on AB different from C is not on the circle. Proof 1. Let D be any point on AB different from C. Connect OD 2. OC is perpendicular to AB (Given) 3. OD>OC (OD is the hypotenuse of the right triangle) 4. D is not on the circle. 5. AB is tangent to the circle (C is the only point on both AB and the circle). Theorem 3 A tangent to a circle is perpendicular to the radius drawn to the point of contact. Given in the figure that ACB is tangent to the circle O at point C. To prove that OC ⊥ AB Proof (We prove by contradiction) 1. Suppose OC is not perpendicular to AB 2. Choose a point D on AB such that OD ⊥ AB (From any point a perpendicular line to a given line can be drawn) 3. Choose point E on AB such that CD = DE and E is on different side of D. O C A O C A B D O C A
  • 3. H3 Mathematics Plane Geometry 3 4. ∠ODC = ∠ODE = 90° 5. OD = OD (Same side) 6. ∆ ODC ≅ ∆ ODE (S.A.S.) 7. OC = OE (Corresponding sides of congruent triangles are equal) 8. E is on O. (by 7) 9. AB intersects the circle at two different points 10. So AB is not tangential to the circle. The contradiction shows that OC ⊥ AB In-Class Exercise 1 Suppose in the given figure that O is the center of the circle, AB and DC are tangents to the circle at B and C respectively. Show that AC = DB. Proof 1. ∠BOA = ∠COD (Vertically opposite angle) 2. OB = OC (radius of the circle) 3. ∠OBA = ∠OCD = 90° (Tangent is perpendicular to radius) 4. ∆ABO ≅ ∆DCO (A.S.A.) 5. AO = DO (Corresponding sides of congruent triangles are equal) 6. AO + OC = DO + OB (OB = OC = radius of the same circle) 7. AC = DB Theorem 4 (Tangent-Chord Theorem) The angle between a tangent and a chord meeting the tangent at the point of contact is equal to the inscribed angle on opposite side of the chord. Given that AB is tangent to the circle at C, CD is a chord, E is on the circle. To prove ∠ DEC =∠ DCB O C A B D E C O B A ED O A D B C
  • 4. H3 Mathematics Plane Geometry 4 Proof 1. Draw the chord FD // AB. 2. Let G be the intersection of ray CO and FD. 3. FD is perpendicular to CG (AB is perpendicular to CG) 4. ∠ DFC =∠ DEC (Intercept the same arc) 5. OF = OD (Radius of the same circle) 6. ∠ OFG =∠ ODG ( ∆ OFD is isosceles) 7. ∠ FGO = ∠ DGO = 90° (by 3) 8. ∆ OFG ≅ ∆ ODG (by 5, 6, 7 and A.S.A) 9. FG = GD (by 8) 10. ∠ FGC =∠ DGC = 90o (by 3) 11. GC = GC 12. ∆ CFG ≅ ∆ CDG (by 9, 10, 11 and S.A.S.) 13. ∠ DFC =∠ FDC (corresponding angles of congruent triangles) 14. ∠ FDC =∠ DCB (alternative interior angles of parallel lines) 15. ∠ DCB =∠ DEC (by 4, 13, 14) In-Class-Exercise 2 Given in the figure that ACB is a tangent of the circle, ∠ ECD = 75o and ∠ DCB = 45o . Find ∠ EDC. Solution ∠EDC = 180° - ∠ DEC - ∠ ECD = 180° - 45° - 75° (Tangent Chord Theorem) = 60° Remark The Tangent-chord theorem can be deduced informally from the fact that inscribed angles that intercept the same arc equal. Imagine the point E in the above figure moves continuously to C along the circle. Then DE approaches CD and EC tends to be parallel to AB. Thus the measurement of ∠ DEC approaches to the measurement of ∠ DCB. F O C A B D E G C A B D E C B A E D
  • 5. H3 Mathematics Plane Geometry 5 Example 1 Suppose in the figure line AB is tangential to the circle at C, DE //AB. Prove (EC)(CG) = (CD)(CF). Proof 1. ∠GFC = ∠GCB (Tangent Chord Theorem) 2. ∠ECB = ∠DEC (Alternate angles) 3. ∠GFC = ∠DEC (by 1 & 2) 4. ∠DCE = ∠GCF (Same angle) 5. ∆DCE ~ ∆GFC (A.A.) 6. CG CD FC EC = (by 5) 7. (EC)(CG) = (CD)(CF) (by 6) Remark From example 1 we can create the following new problems. Can you see why? (1) Suppose in the figure line AB tangents the circle at C, FE //AB. Prove (EC)(CG) = (CF) 2 . D coincides with F, hence CD = CF (2) Suppose in the figure GK is a diameter, FJ is perpendicular to GK. Prove (FG) 2 = (GJ)(GH). FJ is perpendicular to the GK and GK is perpendicular to the Tangent at G, hence FJ is parallel to the tangent at G. Can you create any more new questions? E C BA F G D E C A B F G J G F H K
  • 6. H3 Mathematics Plane Geometry 6 4.3 Intersecting Chord Theorem, Tangent-secant theorem Theorem 5 (Intersecting Chord Theorem) If two chords of a circle intersect in the interior of a circle, thus determine two segments in each chord, the product of the lengths of the segments of one chord equals the product of the lengths of the segments of the other chord. Given that AB and CD are two chords intersecting at P. To prove that (AP)(BP) = (CP)(DP). The plan is to use properties of similar triangles. Proof 1. Connect A and D, B and C. 2. ∠ ADP =∠ CBP (They intercept the same arc) 3. ∠ DAP =∠ BCP (As in 2) 4. ∆ ADP ~∆ CPB (A.A.) 5. AP: CP = DP: BP (by 4) 6. (AP)(BP) = (DP)(CP) (by 5) In-Class-Activity If two intersecting chords bisect each other, what can you say about their relationship? Answer: They are the diameters of the circle Definition 2 A secant of a circle is a line drawn from a point outside the circle that intersects the circle at two points. For example, the line AB is a secant of the circle. A secant is not a tangent. Theorem 6 If from a point outside a circle two secants are drawn, the product of one secant and its external segment is equal to the product of the other secant and its external segment. Given that CBA and CDE are secants from the point C outside the circle. To prove that (CB)(CA) = (CD)(CE). The plan is to show∆ ADC ~∆ EBC P C D A B A B D B C A E P A B D C
  • 7. H3 Mathematics Plane Geometry 7 Proof 1. ∠CEB = ∠DAB (They intercept the same arc) 2. ∠DCA = ∠DCA (Same angle) 3. ∆ECB ~ ∆ACD (A.A.) 4. EC AC BC CD = (Ratio of corresponding sides of similar triangles) 5. (CB)(CA) = (CD)(CE) (by 4) Theorem 7 (Tangent –Secant Theorem) If from a point outside a circle a secant and a tangent are drawn, the secant and its external segment is equal to the square of the tangent. Given tangent AB and secant ACD are from an external point A. To prove that (AB)2 = (AC)(AD). The plan is to show that ∆ ACB ~ ∆ ABD. Proof 1. ∠ CAB =∠ BAD (same angle) 2. ∠ CBA =∠ BDA (By Tangent-chord Theorem) 3. ∆ ACB ~ ∆ ABD (A.A) 4. AB: AD = AC: AB (by 3) 5. (AB)2 = (AC)(AD Example 2 Suppose in the figure the arc DM equals arc ME and BC tangents the circle at M. Prove DB: BM = EC: CM. Proof 1. AM is the bisector of ∠BAC (Given) 2. BA: CA = BM: CM (Tutorial 3 qn 6) 3. BA: BM = CA: CM (by 2) 4. BM: BD = BA: BM and CM: CE = CA: CM (Tangent-Secant Theorem) 5. BM: BD = CM: CE (by 3 & 4) 6. DB: BM = EC: CM (by 5) C B A D C A D B ED M A CB B C A E D
  • 8. H3 Mathematics Plane Geometry 8 Summary • Main terms Inscribed angle, chord, radius, diameter, tangent, secant • Main results Tangent-Chord Theorem Intersecting chord Theorem Tangent-secant Theorem • Useful facts . An inscribed angle has half as many degrees as the intercepted arc. . An angle inscribed in a semicircle is a right angle.