**Central Angles and Inscribed Angles**

The angle formed at the centre of a
circle is called a **central angle** and the angle formed at the circumference of a
circle is called a **circumference angle** or **inscribed angle**.

To understand and prove the **circle
theorems on central angles and inscribed angles**, we use the following axioms on
arcs and angles subtended by them:

1.
Degree measurement of a full circle arc is equivalent to 360°.

2.
Central angle of a circle is equal to the degree measurement of
its opposite arc.

3.
Inscribed angle of a circle is equal to the half of the degree
measurement of its opposite arc.

### Theorems:

**1. ****Central angle of a
circle is double of inscribed angle standing on the same arc.**

**2. ****The angles at the
circumference of a circle standing on the same arc are equal.**

**3. ****Inscribed angle on a
semi-circle is a right angle.**

### Proofs:

**1. ****Central angle of a circle is double of inscribed angle standing on the same arc.**

__Given__**:** O is the centre of
a circle. ∠AOB is a central angle and ∠ACB is
a inscribed angle standing on same arc AB.

__To Prove__**:** ∠AOB = 2∠ACB

__Proof__**:**

__Statements____Reasons__

1.
∠AOB = Arc AB ----> Central angle is equal to its opposite
arc.

2.
∠ACB = ½ Arc AB ---->
Inscribed angle is equal to the half of its opposite arc.

3.
2∠ACB = Arc AB -----> From statements 2.

4. ∠AOB = 2∠ACB -----> From statements 1 and 2.

**Proved.**

**2. ****The angles at the circumference of a circle standing on the same arc are equal.**

__Given__**:** O is the centre of
a circle. ∠ABC and ∠ADC are the inscribed angles standing on same
arc AC.

__To Prove__**:** ∠ABC = ∠ADC

__Proof__**:**

__Statements____Reasons__

1.
∠ABC = ½ Arc AC ----> Inscribed angle is equal to the half of
its opposite arc.

2.
∠ADC = ½ Arc AC ----> Inscribed angle is equal to the half of its opposite
arc.

3.
∠AOB = 2∠ACB -----> From statements 1 and 2.

**Proved. **

**3. ****Inscribed angle on a semi-circle is a right angle.**

__Given__**:** O is the centre of
a circle. AB is a diameter.

__To Prove__**:** ∠ACB = 90°

__Proof__**:**

__Statements____Reasons__

1.
∠AOB = 180° ------> Being ∠AOB a straight angle.

2.
∠ACB = ½ ∠AOB ---> Inscribed angle is equal to the half of its opposite
arc.

3.
∠ACB = ½ × 180° ---> From statements 1 and 2.

4.
∠ACB = 90° -------> From statement 3.

**Proved.**

**Worked Out Examples:**

**Example 1:** Find the value of x and y in the
given figure.

*Solution:** *

*From the figure,*

*∠**OCA
= **∠**OAC
= 50° [Being OA = OC, Radii of same circle]*

*y + 50° + 50° = 180° [Sum of the
angels of DOAC]*

*or, y
+ 100° = 180°*

*or, y
= 180° - 100°*

*or, y
= 80°*

*Now,*

*
x = ½ of y [Inscribed angle is
half of central angle]*

* = ½ × 80°*

* =
40° *

* *

*Example 2:**
Find the value of y in the given figure.*

*Solution:** *

*From the figure,*

*∠**ABC = 90°
[An angle formed on semi-circle]*

*y + 50° + 90° = 180° [Sum of the
angles of DABC]*

*or, y
+ 140° = 180°*

*or, y
= 180° - 140°*

*
= 40° *

* *

*Example 3:**
Find the value of z in the given figure.*

*Solution:** *

*From the figure,*

*Reflex **∠**AOC = 2**∠**ABC [Central angle is double of
inscribed angle on the same arc]*

*
= 2 × 130°*

*
= 260°*

* *

*Now,*

* *

*z + reflex **∠**AOC = 360° [Angles at one complete
rotation]*

*or, z
+ 260° = 360°*

*or, z
= 360° - 260°*

*or, z
= 100°*

* *

*Example 4:**
What is the size of ∠CBX in the given figure?*

*Solution:** *

*From the figure,*

*∠**ADB
= 32° and **∠**BXC
= 85°*

*∠**ACB
= **∠**ADB = 32°
[Inscribed angles on same arc]*

*Now, *

*∠**CBX
+ **∠**ACB
+ **∠**BXC =
180° [Sum of angles of DBXC]*

*or, **∠**CBX + 32° + 85° = 180°*

*or, **∠**CBX + 117°= 180°*

*or, **∠**CBX = 180° - 117°*

*or, **∠**CBX = 63°*

* *

*Example 5:**
In the given diagram, PQR is an isosceles triangle with PQ = PR and a circle is
drawn with PQ as a diameter. Prove that QS = RS.*

*Solution:*

__Given__*:** In Î”PQR, PQ = PR. O is centre and PQ is a
diameter of circle.*

__To Prove__*:** QS = RS.*

__Construction__*:** Points P and S are joined.*

__Proof__*:*

__Statements__* Reasons
*

1. *∠**PSQ = 90° ----> Being **∠**PSQ is in
semi-circle with diameter PQ*

2.
*In Î”PSQ and Î”PSR*

i. *∠**PSQ = **∠**PSR (R) ----> Each of 90°
(Statement 1)*

ii. *PQ = PR (H) -----> By given*

iii. *PS = PS (S) -----> Common side.*

3.
*Î”PSQ **≅** Î”PSR -------> By RHS axiom*

4.
*QS = RS ------> Corresponding
sides of **≅** triangles.*

*Proved.*

* *

*Example 6:**
In the given figure, O is the centre of the circle, PQ the diameter and AO⊥PQ, prove that ∠PBR = ∠PAO.*

*Solution:*

__Given__*:** PQ is the diameter of a circle with centre at O and
AP**⊥**PQ.*

__To Prove__*:** **∠**PBR = **∠**PAO.*

__Construction__*:** Points R and Q are joined.*

__Proof__*:*

*
Statements Reasons
*

1. *∠**PRQ = 90° ----> Being **∠**PRQ is in
semi-circle with diameter PQ*

2.
*In Î”PQR and Î”PAO*

a)
*∠**PRQ = **∠**PAO ----> Being AP**⊥**PQ, and from statement 1.*

b)
*∠**QPR = **∠**APO ----> Common angle.*

c)
*∠**PQR **∠**PAO ---->
Remaining angles.*

3.
*∠**PQR = **∠**PBR -----> Being inscribed angles on same arc PR.*

4.
*∠**PBR = **∠**PAO ----> From statements 1(c) and 3.**
*

*Proved.*

*Example 7:**
In the given figure, O is the centre of the circle, AB the diameter and arc BC
= arc CD. Prove that AD || OC.*

*Solution:*

__Given__*:** O is the centre of circle. AB is the diameter. Arc
BC = arc CD*

__To Prove__*:** AD || OC.*

__Proof__*:*

__Statements__* Reasons
*

1. *∠**BAD = ½ of arc BCD ---> Inscribed angle is equal to the half of its
opposite arc*

2.
*∠**BAD = ½ of arc (BC + CD) ---> whole
part axiom from statement 1*

3.
*∠**BAD = ½ of arc (BC + BC) --->
Being arc BC = arc CD (given).*

4.
*∠**BAD = **½ × arc 2BC ---> From statements 3.*

5.
*∠**BAD = arc BC ---> From statement 4.*

6.
*∠**BOC = arc BC ---> Central angle is
equal to its opposite arc.*

7.
*∠**BAD = **∠**BOC ---> From statements 5 and 6.*

8.
*AD // OC ---> Being corresponding
angles equal (statement 7)*

*Proved.*

* *

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