c20Arc and it’s types:

  • An arc is any part of the circumference.
  • A chord divides the circumference of a circle in two parts or two arcs.
  • The arc which is less than the semi-circle is called minor arc while the one that is larger then than the semi-circle is called major arc.
  • A segment is the part of the circle bounded by an arc and a chord.

Theorem 9: The angle which an arc of a circle subtends at the center is double that which it subtends on any point of the part of the circumference.c210

Given: O  is the center of the given circle. \widehat{APB}   subtends \angle AOB  at the center and \angle ACB  at a point C  on the circumference.

To Prove: \angle AOB = 2 \angle ACB 

Proof: In \triangle AOC  ,

OA = OC  (Radius of the same circle)

Therefore \angle OAC = \angle OCA  (angle opposite to the equal sides of a triangle are equal)

\angle AOD = \angle OAC + \angle OCA  (exterior angle of a \triangle = sum of the opposite interior angles)

= \angle OCA + \angle OCA  41

= 2 \angle OCA 

Similarly in \triangle BOC, \angle BOD = 2 \angle OCB 

Hence \angle AOB = \angle AOD + \angle BOD 

= 2 \angle OCA   + 2 \angle OCB 

= 2 ( \angle OCA   +  \angle OCB) 

= 2 \angle ACB  .

Hence Proved.

Similarly, if you look at the following figures you will see that:


i) \angle AOB = 2 \angle APB

ii) Reflex \ \angle AOB = 2 \angle APB

Theorem 10: Angles in the same segment of a circle are equal.

Given: A center with center O  . \angle ACB \ and \ \angle ADB  are in the same segment of the circle.c27

To Prove: \angle ACB = \angle ADB 

Proof:  From Theorem 9, we know that

\angle AOB = 2 \angle ACB   (angle at the center is twice the angle at the remaining circumference)

Similarly \angle AOB = 2 \angle ADB   (angle at the center is twice the angle at the remaining circumference)

This implies that \angle ACB = \angle ADB  .

Hence Proved.

Similarly, in the adjoining figurec26

i) \angle DAB =  \angle DCB (Angles in the same segment)

ii) \angle ADC =  \angle ABC   (Angles in the same segment)


Theorem 11: The angle in a semi circle is a right angle.

Given: A circle with center O . AB is the diameter and \angle ACB is the angle of semi- circle.c25

To Prove: \angle ACB = 90^o

Proof: \angle AOB = 2 \angle ACB  (angle at the center is twice the angle at the remaining circumference)

Given \angle  AOB = 180^o (straight line)

\Rightarrow 180^o=2 \angle ACB

\Rightarrow \angle ACB = 90^o .

Hence Proved.


Cyclic Properties

Theorem 12: The opposite angles of a cyclic quadrilateral inscribed in a circle are supplementary.c24

Given: A circle with center O . A quadrilateral ABCD inscribed in a circle.

To Prove: \angle ABC + \angle ADC = 180^o \ and \  \angle BAD + \angle BCD = 180^o

Proof:  \angle AOC = 2 \angle ADC (angle at the center is twice the angle at the remaining circumference)

\Rightarrow  \angle  ADC = \frac{1}{2} \angle  AOC

Similarly \angle  ABC = \frac{1}{2}  \ reflex \  \angle  AOC

Adding the two

\angle ABC + \angle  ADC = \frac{1}{2} \angle  AOC + \frac{1}{2}  reflex \angle  AOC

= \frac{1}{2} (\angle  AOC + reflex \  \angle  AOC) = \frac{1}{2} \times 360^o = 180^o

(Reflex  \angle AOC +  \angle AOC = 360^o)

Similarly we can prove that  \angle BAD +  \angle BCD = 180^o .

Hence Proved.

Note: c23

  • If the opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic
  • The angle in the major segment is acute and the angle in the minor segment is obtuse. i.e. \angle ABC < 90^o while \angle ADC > 90^o


Theorem 13: The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.c22

Given: Circle with center O. ABCD is the cyclic quadrilateral.

To Prove: \angle CBE = \angle ADC

Proof:   \angle ABC + \angle CBE = 180^o (the two angles are supplementary)

\angle ABC + \angle ADC = 180^o (Opposite angles of a a cyclic quadrilateral are supplementary…Theorem 12)

Therefore \angle ABC +  \angle  CBE =  \angle  ABC +  \angle  ADC 

\Rightarrow  \angle CBE =  \angle ADC

Hence Proved.

Note: Take a look at the adjoining figure. You will see thatc21x

i) \angle PAB = \angle BCD

ii) \angle QBC = \angle ADC

iii) \angle RCD = \angle BAD