 Arc 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. 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) $= \angle OCA + \angle OCA$ $= 2 \angle OCA$

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. 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 figure 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. 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. 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) $\displaystyle \Rightarrow \angle ADC = \frac{1}{2} \angle AOC$ $\displaystyle \text{ Similarly } \angle ABC = \frac{1}{2} \text{ reflex } \angle AOC$ $\displaystyle \angle ABC + \angle ADC = \frac{1}{2} \angle AOC + \frac{1}{2} \text{ reflex } \angle AOC$ $\displaystyle = \frac{1}{2} (\angle AOC + \text{ reflex } \angle AOC) = \frac{1}{2} \times 360^o = 180^o$ $\displaystyle (\text{ Reflex } \angle AOC + \angle AOC = 360^o)$

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

Hence Proved.

Note: • 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. 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 that i) $\angle PAB = \angle BCD$

ii) $\angle QBC = \angle ADC$

iii) $\angle RCD = \angle BAD$