## Circle Theorems on 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 the inscribed angle standing on 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 the inscribed angle standing on 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 = 2ACB

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.

Or, 2ACB = Arc AB

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

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

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

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 = 2ACB ---------> 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.

Look at the following worked out examples of geometrical problems on circle:

### Workout 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°

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 = 2ABC [Central angle is double of inscribed angle on the same arc]

= 2 × 130°

= 260°

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: Here, ADB = 32° and BXC = 85°

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

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 DPQR, PQ = PR. O is centre and PQ is a diameter of circle.

To Prove: QS = RS.

Construction: Points P and S are joined.

Proof:

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

2.    In DPSQ and DPSR

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

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

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

3.    DPSQ DPSR -------------> 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 AOPQ, prove that PBR = PAO.

Solution:

Given: PQ is the diameter of a circle with centre at O and APPQ.

To Prove: PBR = PAO.

Construction: Points R and Q are joined.

Proof:

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

2.    In DPQR and DPAO

i.        PRQ = PAO ------> Being APPQ, and from statement 1.

ii.       QPR = APO -------> Common angle.

iii.     PQR PAO -------> Remaining angles.

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

4.    PBR = PAO -------------> From statements 1(iii) 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

Proof:

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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