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 OB=OD AB=CD (Radii of the semi circle) (Radii of the semi circle) (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 OB = OD
AOB  COD

and (being radii of the same circles) (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
2  3

(Radii of the same circle) (Since Radii of same circle make isosceles triangle) (External angle is equal to sum or remote interior angles) ( Since 2  3 ) … (1) … (2)

But 1  2  3 or 1  3  3 or 1  23 Similarly 6  24

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
2  3

(Radii of the same circle) (Base angle of an isosceles triangle) (Exterior angle is equal to sum of remote interior angle) ( Since 2  3 ) … (1) … (2)

But 1  2  3 or 1  3  3 or 1  23 Similarly 6  24

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
3  2 1  2  3

(Radii of the same circle)

(Exterior angle of a triangle is equal to sum of remote interior angles) ( Since 2  3 ) … (1) … (2)

or 1  3  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

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
2  1

(Radii of the same circle) (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 or 3  22 Similarly, 4  25 Adding (1) and (2), we get
3  4  22  25  2(2  5)

( Since 1  2 ) … (1) … (2)

or POQ  2PRQ Case II See figure (ii) Adding (1) and (2), we get

3  4  22  25

or180o  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 Therefore ADB  AD ' B

(Angles in the same segment are equal)

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 BAC  BDE

(Angles in the same segment) (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 Therefore D  B  180o In quadrilateral ABCD

(Angle Sum Property of Triangles)

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.