Theorems On Arcs And Angles Subtended By Them

Theorems on Arcs and Angles Subtended by them Theorem: Two arcs of a circle are congruent if the angles subtended by them at the centre are equal. Given: AB and PQ are two arcs of a circles with centers C and ∠ACB = ∠PCQ To prove: arc AB= arc PQ Construction: Draw CM to bisect ∠BCP

Proof: Fold the circle about CM such that CP coincides with CB. Since ∠ACB = ∠PCQ , therefore sector PCQ coincide with sector BCA so that CP falls on CB and CQ falls on CA. Therefore arc AB = arc QP Theorem: (Converse of Previous Theorem) The angles subtended at the centre of a circle by two equal chords are equal Given: AB and PQ are two equal arcs of a circle. To prove: ∠ACB = ∠PCQ Construction: Draw CM bisect ∠BCP

Proof: Fold the circle about CM such that CP coincides with CB. Since arc AB = arc PQ, therefore, when P falls on B, Q will fall on A. Thus, sector PCQ will coincide with sector BCA Hence, ∠ACB = ∠PCQ

Theorem: In two equal circles if two arcs subtend equal angles at the centres, the arcs are equal. Given: Two equal circles with centers C and O and ∠ACB = ∠POQ . To prove: arc AB = arc PQ Proof: Apply circle ABD on circle PQR such that centre C falls on centre O and CA falls along OP.

Since CA = OP

Therefore A falls on P. Also Since ∠ACB = ∠POQ

Therefore CB falls along OB. Since CB = OQ Therefore B falls on Q.

Therefore arc AB coincides with arc PQ Hence arc AB = arc PQ Theorem: (Converse of Previous Theorem) In two equal circles, if two arcs are equal than they subtend equal angles at the centers Given: In two equal circles with centers C and O. arc AB = arc PQ To prove: ∠ACB = ∠POQ

Proof: Apply circles ABD on circle PQR such that centre C falls on centre O and CA falls along OP. Since arc AB = arc PQ, therefore B falls on Q and CB coincides with OQ. Now, CA coincides with OP and CB coincides with OQ.

Therefore ∠ACB = ∠POQ Theorem: Equal chords of circle (or of congruent circles) subtend equal angles at the centre (at the corresponding centers) Given: AB and CD are two equal chords of a circle with centre O. To prove: ∠AOB = ∠COD

Proof: In triangles Triangle AOB and Triangle COD OA=OC

(Radii of the semi circle)

OB=OD

(Radii of the semi circle)

AB=CD

(Given)

Therefore Triangle AOB ≅ Triangle COD (SSS criterion of congruency) Thus, ∠AOB = ∠COD (cpctc) (Proof for congruent circles is similar) Theorem: If the angles subtended by two chords of a circle (or congruent circles) at the centre (or corresponding centers) are equal, then the chords are equal. Given: Two chords AB and CD of a circle with centre O such that ∠AOB = ∠COD To prove: AB=CD

Proof: In triangles Triangle AOB and Triangle COD OA = OC

and

OB = OD

(being radii of the same circles)

∠AOB = ∠COD

(given)

Therefore Triangle AOB ≅ Triangle COD (SAS criterion of congruency) Hence AB = CD (cpctc) Proof for congruent circles is similar Theorem: The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle. Or The degree measure of an arc of a circle is twice the angle subtended by it at any point of the alternate segment of the circle with respect to the arc. Given: A circle with centre O. A chord AB subtends an angle at the centre making ∠AOB . It also subtends an angle with circle at C making an angle ∠ACB . To prove: ∠AOB = 2∠ACB

Proof: Case I: When ∠AOB is an acute angle In Triangle AOC , OA = OC

(Radii of the same circle)

∠2 = ∠3

(Since Radii of same circle make isosceles triangle)

But ∠1 = ∠2 + ∠3

(External angle is equal to sum or remote interior angles)

or ∠1 = ∠3 + ∠3

( Since ∠2 = ∠3 )

or ∠1 = 2∠3

… (1)

Similarly ∠6 = 2∠4

… (2)

Adding (1) and (2), we have

∠1 + ∠6 = 2∠3 + 2∠4 = 2(∠3 + ∠4) or ∠AOB = 2∠ACB Case II: When ∠AOB is straight angle

In this case ∠AOB = 180o AO = OC

(Radii of the same circle)

∠2 = ∠3

(Base angle of an isosceles triangle)

But ∠1 = ∠2 + ∠3

(Exterior angle is equal to sum of remote interior angle)

or ∠1 = ∠3 + ∠3

( Since ∠2 = ∠3 )

or ∠1 = 2∠3

… (1)

Similarly ∠6 = 2∠4

… (2)

Adding (1) and (2), we get ∠1 + ∠6 = 2∠3 + 2∠4

or ∠1 + ∠6 = 2(∠3 + ∠4)

Therefore ∠AOB = 2∠ACB But ∠AOB = 180o

Therefore 2∠ACB = 180o or ∠ACB =

180o = 90o which is true as an angle in a semi-circle is a right angle. 2

Case III: When ∠AOB is obtuse reflex angle In Triangle OAC , we have OA = OC

(Radii of the same circle)

∠3 = ∠2 ∠1 = ∠2 + ∠3

(Exterior angle of a triangle is equal to sum of remote interior angles)

or ∠1 = ∠3 + ∠3

( Since ∠2 = ∠3 )

or ∠1 = 2∠3 Similarly in Triangle BOC , we have ∠6 = 2∠4

Adding (1) and (2), we get ∠1 + ∠6 = 2∠3 + 2∠4 = 2(∠3 + ∠4) or ∠AOB = 2∠ACB

… (1) … (2)

Theorem: The degree measure of an arc of a circle is twice the angle subtended by it at any point of the alternate segment of the circle with respect to the arc. Given: An arc PQ of circle C ( O, r ) with a point R in arc PQ other than P or Q. To prove: Measure of arc PQ = 2 ∠PRQ Proof:

Case-I When arc PQ is minor arc. See figure (i) In Triangle PQR, OP = OR

(Radii of the same circle)

∠2 = ∠1

(Base angles of an isosceles triangle)

Now ∠3 = ∠1 + ∠2 (Exterior angle is equal to sum of the remote interior angles) or ∠3 = ∠2 + ∠2

( Since ∠1 = ∠2 )

or ∠3 = 2∠2

… (1)

Similarly, ∠4 = 2∠5

… (2)

Adding (1) and (2), we get

∠3 + ∠4 = 2∠2 + 2∠5 = 2(∠2 + ∠5) or ∠POQ = 2∠PRQ Case II See figure (ii) Adding (1) and (2), we get

∠3 + ∠4 = 2∠2 + 2∠5

or 180o = 2(∠2 + ∠5) or ∠POQ = 2∠PRQ Case III:

∠3 + ∠4 = (180o − ∠6) + (180o − ∠7) o 2∠2 + 2∠5 = 360o − (∠6 + ∠7) or 2(∠2 + ∠5) = 360o − 2∠POQ or ∠POQ = 2∠PRQ Hence for all the figures (i), (ii) and (iii) Measure of arc PQ = 2∠PRQ or m (PQ)= 2∠PRQ Theorem: Any two angles in the same segment are equal. Given: A circle whose centre is O. ∠2 and ∠3 are on the same segment

To prove: ∠2 = ∠3 Construction: Join OA and OB Proof: arc AB subtends ∠1 at the centre and ∠3 on the remaining part of it. Therefore ∠1 = 2∠3 or 2∠2 = 2∠3 or ∠2 = ∠3 Theorem: An angle in a semi-circle is a right angle Given: A circle with centre O and AB is the diameter. ∠ACB is an angle in the semi-circle To prove: ∠ACB = 90o Proof: Since the angle subtended by an arc at the centre is double the angle subtended by it at the remaining part of the circumference,

Therefore ∠AOB = 2∠ACB

But ∠AOB = 180o

( Since ∠AOB is a straight angle)

2∠ACB = 180o or ∠ACB =

180o = 90o 2

Theorem: If a line segment joining two points subtends equal angles at two other points lying on the same side of the line segment, the four points are con-cyclic, that is they lie on the same circle.

Given: AB is the line segment and C, D are two points lying on the same side of the line AB such that ∠ACB = ∠ADB To prove: A,B,C, D are concyclic Construction: Draw a circle through three non-collinear points A, B and C Proof: Case I If D lies on the circle through A, B and C, then the required result gets proved. Case-II If possible, suppose D does not lie on the circle drawn through A, B and C. Then the circle will intersect AD or AD produced D ' . Join D ' B . Now, ∠ACB = ∠ADB

(Given)

∠ADB = ∠AD ' B

(Angles in the same segment are equal)

Therefore ∠ADB = ∠AD ' B But this is not possible because an exterior angle of a triangle ( Triangle BDD ' in this case) cannot be equal to its interior opposite angle.

Therefore ∠ADB and ∠AD ' B will be equal only if D and D ' coincide. Therefore D will lie on the circle through A, B and C. Hence points A, B, C and D are con-cyclic. Theorem: The sum of either pair of opposite angles of a cyclic quadrilateral is 180o or the opposite angle o a cyclic quadrilateral are supplementary. Given: ABCD is a cyclic quadrilateral. To prove: (i) ∠A + ∠C = 180o

(ii) ∠B + ∠D = 180o

Construction: Join AC and BD

Proof:

∠ACB = ∠ADE

(Angles in the same segment)

∠BAC = ∠BDE

(Angles in the same segment)

Adding the above equation, we have ∠ACB + ∠BAC = ∠ADE + ∠BDC

or ∠ACB + ∠BAC = ∠ADC Now adding ∠ABC to both sides we get ∠ACB + ∠BAC + ∠ABC = ∠ADC + ∠ABC

or 180o = ∠ADC + ∠ABC

(Angle Sum Property of Triangles)

Therefore ∠D + ∠B = 180o In quadrilateral ABCD

∠A + ∠B + ∠C + ∠D = 360o (Sum of angles of a quadrilateral is 360o ) or (∠A + ∠C ) + (∠B + ∠D) = 360o or (∠A + ∠C ) + 180o = 360o ( Since ∠D + ∠B = 180o , proved above) or (∠A + ∠C ) = 360o − 180o or (∠A + ∠C ) = 180o Hence (∠A + ∠C ) = 180o and ∠D + ∠B = 180o Theorem: (Converse of Previous Theorem): If the sum of any pair of opposite angles of a quadrilateral is 180o , then the quadrilateral is cyclic.

Proof: Quadrilateral ABCD is which (∠A + ∠C ) = 180o To prove: ABCD is a cyclic quadrilateral Proof: Suppose, If possible, that ABCD is not cycle. Now draw a circle passing through three non-collinear points A,B and C. Let the circle meets DC or DC ' produced C ' . Join C ' B . Since ABCD is a cyclic quadrilateral,

∠DC ' B = ∠C Now, an exterior angle of a triangle cannot be equal to its remote interior angle. This is ridicule Thus the circle passing through D, A, B passes through C also. Hence ABCD is a cyclic quadrilateral. ******************************************