Chapter 17: Circles - Vardaan Master Notes

Created by Team Vardaan with Powered by

Official ICSE Syllabus & Chapter Scope

Prescribed Syllabus: Circles – Chord properties, Angle properties, Cyclic properties, Inscribed and Circumscribed circles, and Angle-Arc relationships.

Scope of Syllabus:

Chord Properties: Perpendicular from centre to a chord bisects the chord; Equal chords are equidistant from centre; Equal chords subtend equal angles at centre.
Arc & Angle Properties: Angle at centre is double the angle at remaining circumference ($\angle AOB = 2\angle ACB$); Angles in the same segment of a circle are equal; Angle in a semi-circle is a right angle ($90^\circ$).
Cyclic Properties: Opposite angles of a cyclic quadrilateral are supplementary ($\angle A + \angle C = 180^\circ$); Exterior angle of a cyclic quadrilateral is equal to the interior opposite angle; Properties of cyclic parallelograms, rhombuses, and trapezia.
Arc & Angle Proportionality: Length of arcs proportional to central angles; Inscribed regular polygons.
Solved Masterclass: Comprehensive geometric proofs and calculation numericals with formal Statement-Reason proofs and complete vector diagrams.

PART (A): Fundamental Concepts & Chord Properties

1. Circle Terminology and Fundamental Definitions

Definitions & Locus Concept

Circle: A circle is the closed plane curve obtained by joining all points in a plane that are at a constant distance (radius) from a fixed point (centre).
Locus Definition: A circle is the locus of a point which moves in a plane such that its distance from a fixed point in the same plane remains constant.
Radius ($r$): The constant distance from the centre $O$ to any point on the circle.
Circumference ($C$): The perimeter or boundary length of the circle ($C = 2\pi r$).
Chord: A line segment joining any two points on the circumference of a circle.
Diameter ($d$): A chord passing through the centre of the circle ($d = 2r$). It is the longest chord of the circle.
Secant: A straight line intersecting a circle at two distinct points and extending outside it.

Basic Circle Elements: Centre (O), Radius (OC), Diameter (AB), and Chord (PQ).

2. Types of Circles

Type of Circle	Definition & Characteristic Property
Concentric Circles	Two or more circles having the same centre $O$ but different radii ($r_1 \neq r_2$).
Congruent (Equal) Circles	Circles having equal radii ($r_1 = r_2$). They have identical circumference and area.
Circumscribed Circle	A circle passing through all vertices of a polygon. Its centre is called the circumcentre.
Inscribed Circle (In-circle)	A circle touching all sides of a polygon internally. Its centre is the incentre (intersection of internal angle bisectors).

3. Fundamental Theorems on Chords

THEOREM: PERPENDICULAR BISECTOR OF CHORD

Theorem 1: The straight line drawn from the centre of a circle to bisect a chord (which is not a diameter) is perpendicular to the chord.

In a circle with centre $O$, if $M$ is the mid-point of chord $AB$ ($AM = MB$):

$$ OM \perp AB \implies \angle OMA = \angle OMB = 90^\circ $$

Theorem 2 (Converse): The perpendicular drawn from the centre of a circle to a chord bisects the chord.

$$ OM \perp AB \implies AM = MB = \frac{1}{2}AB $$

Theorem 1 & 2: OM ⊥ AB ⟺ AM = MB = AB/2.

⭐ PYTHAGOREAN RELATION IN CHORD PROBLEMS

In $\Delta OMA$ right-angled at $M$, where $r = OA$ (radius), $d = OM$ (perpendicular distance from centre), and $AM = \frac{1}{2}AB$:

$$ r^2 = OM^2 + AM^2 \quad \implies \quad r = \sqrt{d^2 + \left(\frac{AB}{2}\right)^2} $$

4. Chord Length vs Distance from Centre

Distance & Length Inverse Relation

Greater the chord, smaller is its perpendicular distance from the centre:
If $\text{chord } AB > \text{chord } CD \implies OM < ON$ (where $OM \perp AB$ and $ON \perp CD$).
Converse: If $OM < ON \implies AB > CD$.
Three Points Rule: There is one and only one circle which passes through three given non-collinear points.

5. Equal Chords and Distance Properties

EQUAL CHORDS EQUIDISTANCE THEOREM

Theorem 3: Equal chords of a circle are equidistant from the centre. $$ AB = CD \implies OM = ON $$
Theorem 4 (Converse): Chords of a circle that are equidistant from the centre are equal in length. $$ OM = ON \implies AB = CD $$

Theorem 3 & 4: AB = CD ⟺ OM = ON.

6. Classroom Practice Problems (Chord Theorems & Distance from Centre)

✍ PRACTICE PROBLEM A1: CHORD LENGTH, RADIUS & PYTHAGOREAN RELATION

Q. In a circle of radius $10\text{ cm}$, a chord $AB$ of length $16\text{ cm}$ is drawn. (i) Find the distance of chord $AB$ from the centre $O$. (ii) Find the length of another chord $CD$ in the same circle which is at a distance of $6\text{ cm}$ from the centre.

(i) Find Distance of Chord $AB$ from Centre ($OM$):

Let $OM \perp AB$. By Theorem 2, perpendicular from centre bisects the chord: $$ AM = \frac{1}{2}AB = \frac{16}{2} = 8\text{ cm} $$
In right-angled $\Delta OMA$, by Pythagoras Theorem ($OA = r = 10\text{ cm}$): $$ OM^2 + AM^2 = OA^2 \implies OM^2 + 8^2 = 10^2 $$ $$ OM^2 = 100 - 64 = 36 \implies \mathbf{OM = 6\text{ cm}} $$

(ii) Find Length of Chord $CD$ at Distance $ON = 6\text{ cm}$ from Centre:

In right-angled $\Delta ONC$ ($OC = r = 10\text{ cm}, ON = 6\text{ cm}$): $$ CN^2 = OC^2 - ON^2 = 10^2 - 6^2 = 100 - 36 = 64 \implies CN = 8\text{ cm} $$
Since $ON \perp CD$, $N$ bisects $CD$: $$ CD = 2 \times CN = 2 \times 8 = \mathbf{16\text{ cm}} $$
Note: Chords equidistant from the centre are equal in length ($OM = ON = 6\text{ cm} \implies AB = CD = 16\text{ cm}$).

✍ PRACTICE PROBLEM A2: PARALLEL CHORDS ON OPPOSITE & SAME SIDES OF CENTRE

Q. In a circle of radius $10\text{ cm}$, two parallel chords $AB$ and $CD$ of lengths $12\text{ cm}$ and $16\text{ cm}$ respectively are drawn. Calculate the distance between the two parallel chords when:
(i) They lie on opposite sides of the centre.
(ii) They lie on the same side of the centre.

Step 1: Calculate Distance of Each Chord from Centre $O$:

Draw perpendiculars $OM \perp AB$ and $ON \perp CD$.
For chord $AB = 12\text{ cm} \implies AM = \frac{12}{2} = 6\text{ cm}$. In right $\Delta OMA$: $$ OM = \sqrt{OA^2 - AM^2} = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8\text{ cm} $$
For chord $CD = 16\text{ cm} \implies CN = \frac{16}{2} = 8\text{ cm}$. In right $\Delta ONC$: $$ ON = \sqrt{OC^2 - CN^2} = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6\text{ cm} $$

Case (i): When Chords lie on OPPOSITE SIDES of Centre: $$ \text{Distance between chords } MN = OM + ON = 8\text{ cm} + 6\text{ cm} = \mathbf{14\text{ cm}} $$

Case (ii): When Chords lie on the SAME SIDE of Centre: $$ \text{Distance between chords } MN = OM - ON = 8\text{ cm} - 6\text{ cm} = \mathbf{2\text{ cm}} $$

PART (B): Arcs, Segments & Subtended Angle Theorems

1. Arcs and Segments of a Circle

Arc & Segment Terminology

Arc: A continuous piece or portion of the circumference of a circle.
- Minor Arc: An arc whose length is less than the semi-circle (e.g. arc $APB$).
- Major Arc: An arc whose length is greater than the semi-circle (e.g. arc $AQB$).
- Semi-circle: An arc whose endpoints form a diameter ($180^\circ$ measure).
Segment: The region bounded by a chord and the corresponding arc.
- Minor Segment: Region bounded by a chord and its minor arc.
- Major Segment: Region bounded by a chord and its major arc.

2. Theorem 5: Angle Subtended by an Arc at the Centre

CORE THEOREM: ANGLE AT CENTRE = 2 × ANGLE AT CIRCUMFERENCE

Statement: 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 circumference.

$$ \angle AOB = 2 \angle ACB \quad \iff \quad \angle ACB = \frac{1}{2} \angle AOB $$

Theorem 5: Angle at centre ∠AOB (2θ) is double the angle at circumference ∠ACB (θ).

Formal Geometric Proof of Theorem 5

Given: A circle with centre $O$. Arc $APB$ subtends $\angle AOB$ at the centre and $\angle ACB$ at point $C$ on the remaining circumference.

To Prove: $\angle AOB = 2 \angle ACB$.

Construction: Join $CO$ and produce it to a point $D$.

Statement	Reason
1. In $\Delta AOC$, $OA = OC$	Radii of the same circle.
2. $\therefore \angle OAC = \angle OCA$	Angles opposite to equal sides in a triangle are equal.
3. Ext. $\angle AOD = \angle OAC + \angle OCA = 2\angle OCA$	Exterior angle of a $\Delta$ equals sum of two interior opposite angles.
4. Similarly, in $\Delta BOC$, Ext. $\angle BOD = 2\angle OCB$	Same reasoning applied to $\Delta BOC$ ($OB = OC$).
5. $\angle AOB = \angle AOD + \angle BOD = 2\angle OCA + 2\angle OCB$	Adding equations (3) and (4).
6. $\therefore \angle AOB = 2(\angle OCA + \angle OCB) = \mathbf{2 \angle ACB}$	Hence Proved!

⭐ CASE OF MAJOR ARC (REFLEX ANGLE)

When arc $AB$ is a major arc, the central angle is a reflex angle:

$$ \text{Reflex } \angle AOB = 2 \angle APB $$

3. Theorem 6: Angles in the Same Segment

THEOREM: ANGLES IN SAME SEGMENT ARE EQUAL

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

$$ \angle ACB = \angle ADB $$

Theorem 6: Angles in the same segment ∠ACB = ∠ADB = θ.

Proof of Theorem 6

Given: Angles $\angle ACB$ and $\angle ADB$ in the same segment subtended by chord/arc $AB$.

Proof:

$\angle AOB = 2\angle ACB$ (Angle at centre is double angle at circumference).
$\angle AOB = 2\angle ADB$ (Same theorem applied to $\angle ADB$).
Equating both relations: $$ 2\angle ACB = 2\angle ADB \implies \mathbf{\angle ACB = \angle ADB} $$ Hence Proved!

4. Theorem 7: Angle in a Semi-Circle

THEOREM: ANGLE IN A SEMI-CIRCLE IS 90°

Statement: The angle in a semi-circle is a right angle ($90^\circ$).

If $AB$ is a diameter of the circle, then for any point $C$ on the semi-circle:

$$ \mathbf{\angle ACB = 90^\circ} $$

Theorem 7: Angle in a semi-circle ∠ACB = 90°.

Proof of Theorem 7

Diameter $AOB$ is a straight line passing through centre $O$, so the central angle $\angle AOB = 180^\circ$.
By Theorem 5, the angle subtended at the circumference is half the central angle: $$ \angle AOB = 2 \angle ACB \implies 180^\circ = 2 \angle ACB \implies \mathbf{\angle ACB = 90^\circ} $$ Hence Proved!

5. Angle in Major vs Minor Segment

Acute & Obtuse Segment Rules

Angle in a Major Segment is ACUTE ($< 90^\circ$): The arc is smaller than a semi-circle, so central angle $< 180^\circ \implies \angle < 90^\circ$.
Angle in a Minor Segment is OBTUSE ($> 90^\circ$): The reflex central angle $> 180^\circ \implies \angle > 90^\circ$.

6. Classroom Practice Problems (Angle at Centre, Semi-Circle & Same Segment)

✍ PRACTICE PROBLEM B1: ANGLE AT CENTRE & SUBTENDED ANGLES

Q. In a circle with centre $O$, central angle $\angle AOC = 110^\circ$. $D$ is a point on the major arc and $B$ is a point on the minor arc. Calculate (i) $\angle ADC$, (ii) $\angle ABC$, (iii) $\angle OAC$.

(i) Find $\angle ADC$: By Theorem 5, angle at circumference is half the angle at centre: $$ \angle ADC = \frac{1}{2} \angle AOC = \frac{1}{2} \times 110^\circ = \mathbf{55^\circ} $$

(ii) Find $\angle ABC$: $ABCD$ is a cyclic quadrilateral, so opposite angles are supplementary: $$ \angle ABC + \angle ADC = 180^\circ \implies \angle ABC = 180^\circ - 55^\circ = \mathbf{125^\circ} $$

(iii) Find $\angle OAC$: In $\Delta AOC$, $OA = OC$ (radii of circle): $$ \angle OAC = \angle OCA = \frac{180^\circ - \angle AOC}{2} = \frac{180^\circ - 110^\circ}{2} = \frac{70^\circ}{2} = \mathbf{35^\circ} $$

✍ PRACTICE PROBLEM B2: ANGLES IN THE SAME SEGMENT

Q. In a circle, chord $PQ = PR$ and $\angle PRQ = 70^\circ$. Find $\angle QAR$ where $A$ lies on the major arc.

In $\Delta PQR$: Since $PQ = PR$, angles opposite to equal sides are equal: $$ \angle PQR = \angle PRQ = 70^\circ $$

Third angle $\angle QPR$: $$ \angle QPR = 180^\circ - (70^\circ + 70^\circ) = 180^\circ - 140^\circ = 40^\circ $$

Find $\angle QAR$: $\angle QAR$ and $\angle QPR$ lie in the same segment of chord $QR$: $$ \angle QAR = \angle QPR = \mathbf{40^\circ} $$

✍ PRACTICE PROBLEM B3: ANGLES IN SAME SEGMENT & TRIANGLE SUM

Q. A circle passes through points $A, B, C, D$. If $\angle BAC = 67^\circ$, find $\angle DBC + \angle DCB$.

Angles in Same Segment: Chord $BC$ subtends $\angle BAC$ and $\angle BDC$ at the circumference: $$ \angle BDC = \angle BAC = 67^\circ $$

In $\Delta DBC$: Sum of all three angles is $180^\circ$: $$ \angle BDC + \angle DBC + \angle DCB = 180^\circ $$ $$ 67^\circ + (\angle DBC + \angle DCB) = 180^\circ \implies \mathbf{\angle DBC + \angle DCB = 180^\circ - 67^\circ = 113^\circ} $$

✍ PRACTICE PROBLEM B4: ANGLE IN A SEMI-CIRCLE & DIAMETER

Q. In a circle, $AC$ is a diameter. $AB = BC$ and $\angle AED = 118^\circ$. Calculate (i) $\angle DEC$, (ii) $\angle DAB$.

(i) Angle in a Semi-circle: Join $EC$. Since $AC$ is diameter, by Theorem 7, $\angle AEC = 90^\circ$. $$ \angle DEC = \angle AED - \angle AEC = 118^\circ - 90^\circ = \mathbf{28^\circ} $$

(ii) Find $\angle DAB$:

$\angle DAC = \angle DEC = 28^\circ$ (angles in the same segment of chord $DC$).
In $\Delta ABC$, $\angle ABC = 90^\circ$ (angle in semi-circle) and $AB = BC \implies \angle BAC = \angle BCA = 45^\circ$.
$\therefore \angle DAB = \angle DAC + \angle BAC = 28^\circ + 45^\circ = \mathbf{73^\circ}$.

✍ PRACTICE PROBLEM B5: PARALLEL LINES & ANGLE AT CENTRE

Q. In a circle with centre $O$, $BC \parallel DE$ and diameter $DOE$ is drawn. If $\angle CDE = x^\circ$, find in terms of $x^\circ$ the value of $\angle BAC$.

Alternate Interior Angles: Since $BC \parallel DE$ and $DC$ is the transversal: $$ \angle BCD = \angle CDE = x^\circ $$

Central Angles: Join $OB$ and $OC$. $$ \angle COE = 2 \angle CDE = 2x^\circ \quad \text{and} \quad \angle BOD = 2 \angle BCD = 2x^\circ $$

Straight Line Diameter $DOE$: $$ \angle DOB + \angle BOC + \angle COE = 180^\circ \implies 2x^\circ + \angle BOC + 2x^\circ = 180^\circ $$ $$ \angle BOC = 180^\circ - 4x^\circ $$

Find $\angle BAC$: $$ \angle BAC = \frac{1}{2} \angle BOC = \frac{180^\circ - 4x^\circ}{2} = \mathbf{(90 - 2x)^\circ} $$

PART (C): Cyclic Quadrilaterals & Cyclic Properties

1. Cyclic Quadrilateral & Concyclic Points

Cyclic Quadrilateral: A quadrilateral whose all four vertices lie on the circumference of a circle.

Concyclic Points: Four or more points are called concyclic if a circle can be drawn passing through all of them.

2. Theorem 8: Opposite Angles of a Cyclic Quadrilateral

THEOREM: OPPOSITE ANGLES ARE SUPPLEMENTARY

Statement: The opposite angles of a cyclic quadrilateral (quadrilateral inscribed in a circle) are supplementary ($180^\circ$).

$$ \angle ABC + \angle ADC = 180^\circ \quad \text{and} \quad \angle BAD + \angle BCD = 180^\circ $$

Theorem 8: In cyclic quad ABCD, ∠B + ∠D = 180° and ∠A + ∠C = 180°.

Formal Proof of Theorem 8

Given: Cyclic quadrilateral $ABCD$ inscribed in a circle with centre $O$.

To Prove: $\angle ABC + \angle ADC = 180^\circ$ and $\angle BAD + \angle BCD = 180^\circ$.

Construction: Join $OA$ and $OC$.

Statement	Reason
1. $\angle ADC = \frac{1}{2} \angle AOC$	Angle at centre is double the angle at remaining circumference.
2. $\angle ABC = \frac{1}{2} \text{reflex } \angle AOC$	Major arc $ADC$ subtends reflex $\angle AOC$ at centre.
3. $\angle ABC + \angle ADC = \frac{1}{2}(\text{reflex }\angle AOC + \angle AOC)$	Adding (1) and (2).
4. $\angle ABC + \angle ADC = \frac{1}{2} \times 360^\circ = \mathbf{180^\circ}$	Sum of angles at a point $= 360^\circ$.
5. Since sum of all angles in a quad $= 360^\circ$, $\therefore \mathbf{\angle BAD + \angle BCD = 180^\circ}$	$360^\circ - 180^\circ = 180^\circ$. Hence Proved!

⭐ CONVERSE OF THEOREM 8

If the sum of any pair of opposite angles of a quadrilateral is $180^\circ$, then the quadrilateral is CYCLIC.

3. Theorem 9: Exterior Angle of a Cyclic Quadrilateral

THEOREM: EXT. ANGLE = INTERIOR OPPOSITE ANGLE

Statement: If one side of a cyclic quadrilateral is produced, the exterior angle so formed is equal to the interior opposite angle.

If side $AB$ is produced to $E$:

$$ \mathbf{\text{Ext. } \angle CBE = \angle ADC} $$

Theorem 9: Exterior angle Ext. ∠CBE = Interior opposite angle ∠ADC (θ).

Proof of Theorem 9

$\angle ABC + \text{Ext. } \angle CBE = 180^\circ$ (Linear pair on straight line $ABE$).
$\angle ABC + \angle ADC = 180^\circ$ (Opposite angles of cyclic quadrilateral $ABCD$).
Equating both relations: $$ \angle ABC + \text{Ext. } \angle CBE = \angle ABC + \angle ADC $$ $$ \implies \mathbf{\text{Ext. } \angle CBE = \angle ADC} $$ Hence Proved!

4. Special Geometric Deductions on Cyclic Figures

Geometric Theorem	Proof Summary & Core Relation
A cyclic parallelogram is a RECTANGLE	In parallelogram $ABCD$, $\angle A = \angle C$ (opp. angles equal). In cyclic quad, $\angle A + \angle C = 180^\circ \implies 2\angle A = 180^\circ \implies \angle A = 90^\circ$. A $\\|gm$ with one right angle is a rectangle!
A cyclic rhombus is a SQUARE	A rhombus is a parallelogram with all 4 sides equal. Since cyclic $\\|gm$ is a rectangle, all angles are $90^\circ$. A rhombus with $90^\circ$ angles is a square!
A cyclic trapezium is ISOSCELES	If $AB \parallel DC$ in cyclic quad $ABCD$, co-interior angles $\angle A + \angle D = 180^\circ$. Also cyclic opp. angles $\angle B + \angle D = 180^\circ \implies \angle A = \angle B$. This proves non-parallel sides $AD = BC$ and diagonals $AC = BD$.

5. Classroom Practice Problems (Cyclic Quadrilaterals & Exterior Angles)

✍ PRACTICE PROBLEM C1: CYCLIC OPPOSITE ANGLES & LINEAR EQUATIONS

Q. The opposite angles of a cyclic quadrilateral $ABCD$ are given by $\angle A = (2x + 4)^\circ, \angle B = (x + 10)^\circ, \angle C = (4x - 4)^\circ$, and $\angle D = (5y + 5)^\circ$. Find the values of $x$ and $y$, and hence calculate all four angles of the quadrilateral.

Step 1: Opposite Angles $\angle A$ and $\angle C$ are Supplementary: $$ \angle A + \angle C = 180^\circ \implies (2x + 4) + (4x - 4) = 180^\circ $$ $$ 6x = 180^\circ \implies \mathbf{x = 30^\circ} $$

Step 2: Opposite Angles $\angle B$ and $\angle D$ are Supplementary: $$ \angle B + \angle D = 180^\circ \implies (x + 10) + (5y + 5) = 180^\circ $$ Substitute $x = 30$: $$ (30 + 10) + 5y + 5 = 180^\circ \implies 45 + 5y = 180^\circ \implies 5y = 135^\circ \implies \mathbf{y = 27^\circ} $$

Step 3: Calculate the Four Angles:

$\angle A = 2(30) + 4 = \mathbf{64^\circ}$
$\angle B = 30 + 10 = \mathbf{40^\circ}$
$\angle C = 4(30) - 4 = \mathbf{116^\circ}$
$\angle D = 5(27) + 5 = 135 + 5 = \mathbf{140^\circ}$

✍ PRACTICE PROBLEM C2: CYCLIC QUADRILATERAL & EXTERIOR ANGLES

Q. In a cyclic quadrilateral $ABCD$, $\angle CBQ = 48^\circ$ and $a = 2b$. Calculate the numerical value of $b$.

In $\Delta BCQ$: Exterior angle $\angle DCB = 48^\circ + b$.

In $\Delta APB$: Vertically opposite angle $\angle ABP = \angle CBQ = 48^\circ \implies \angle DAB = 48^\circ + a$.

Cyclic Property: Opposite angles are supplementary: $$ \angle DCB + \angle DAB = 180^\circ \implies (48^\circ + b) + (48^\circ + a) = 180^\circ $$ $$ a + b = 180^\circ - 96^\circ = 84^\circ $$ Substitute $a = 2b$: $$ 2b + b = 84^\circ \implies 3b = 84^\circ \implies \mathbf{b = 28^\circ} $$

✍ PRACTICE PROBLEM C3: PROVING DIAMETER VIA RIGHT ANGLE

Q. In a cyclic quadrilateral $ABCD$, $\angle BAD = 80^\circ$, $\angle ABD = 55^\circ$, and $\angle BDC = 45^\circ$. Find (i) $\angle BCD$, (ii) $\angle ADB$. Hence show that diagonal $AC$ is a diameter of the circle.

(i) Find $\angle BCD$: $\angle BCD = 180^\circ - \angle BAD = 180^\circ - 80^\circ = \mathbf{100^\circ}$.

(ii) In $\Delta ABD$: $\angle ADB = 180^\circ - (80^\circ + 55^\circ) = \mathbf{45^\circ}$.

Show $AC$ is Diameter: $$ \angle ADC = \angle ADB + \angle BDC = 45^\circ + 45^\circ = 90^\circ $$ Since $\angle ADC = 90^\circ$ is subtended on the circumference, chord $AC$ must be a diameter (by Converse of Theorem 7)!

✍ PRACTICE PROBLEM C4: CYCLIC TRAPEZIUM IS ISOSCELES

Q. If two opposite sides of a cyclic quadrilateral are parallel ($AB \parallel DC$), prove that the other two non-parallel sides are equal ($AD = BC$).

Proof: Let $ABCD$ be a cyclic quadrilateral with $AB \parallel DC$.

$\angle ABC + \angle ADC = 180^\circ$ (Opposite angles of cyclic quad are supplementary).
$\angle BAD + \angle ADC = 180^\circ$ (Co-interior angles since $AB \parallel DC$).
$\implies \angle ABC + \angle ADC = \angle BAD + \angle ADC \implies \angle ABC = \angle BAD$.

Congruence of Triangles: In $\Delta BAD$ and $\Delta ABC$:

$AB = AB$ (Common side).
$\angle BAD = \angle ABC$ (Proved above).
$\angle ADB = \angle ACB$ (Angles in same segment of chord $AB$).
$\therefore \Delta BAD \cong \Delta ABC$ by AAS congruence rule.
$\implies \mathbf{AD = BC}$ (CPCTC). Hence Proved!

PART (D): Arc-Angle Proportionality & Inscribed Polygons

1. Relation Between Arcs, Chords, and Central Angles

ARC - ANGLE PROPORTIONALITY

1. In a circle (or congruent circles), equal chords subtend equal angles at the centre, and conversely:

$$ \text{chord } AB = \text{chord } CD \iff \angle AOB = \angle COD \iff \text{arc } AB = \text{arc } CD $$

2. The lengths of two arcs of a circle are in the same ratio as the angles subtended by them at the centre:

$$ \frac{\text{Length of Arc } AB}{\text{Length of Arc } CD} = \frac{\angle AOB}{\angle COD} $$

Arc-Angle Relation: Arc AB / Arc CD = ∠AOB (α) / ∠COD (β).

2. Central Angles of Inscribed Regular Polygons

If a regular polygon of $n$ sides is inscribed in a circle, each side subtends an angle $\theta$ at the centre:

$$ \theta = \frac{360^\circ}{n} $$

Inscribed Regular Polygon	Number of Sides ($n$)	Central Angle Subtended by Each Side ($\theta = 360^\circ/n$)
Equilateral Triangle	$n = 3$	$\angle AOB = \frac{360^\circ}{3} = \mathbf{120^\circ}$
Square	$n = 4$	$\angle AOB = \frac{360^\circ}{4} = \mathbf{90^\circ}$
Regular Pentagon	$n = 5$	$\angle AOB = \frac{360^\circ}{5} = \mathbf{72^\circ}$
Regular Hexagon	$n = 6$	$\angle AOB = \frac{360^\circ}{6} = \mathbf{60^\circ}$
Regular Octagon	$n = 8$	$\angle AOB = \frac{360^\circ}{8} = \mathbf{45^\circ}$

Inscribed Regular Hexagon (n=6): Each side subtends central angle θ = 360°/6 = 60° (side = radius r).

3. Classroom Practice Problems (Arc Ratios & Inscribed Regular Polygons)

✍ PRACTICE PROBLEM D1: RATIO OF ARCS & CENTRAL/CIRCUMFERENCE ANGLES

Q. The lengths of arc $AB$ and arc $BC$ of a circle are in the ratio $3 : 2$. If the angle subtended by arc $AB$ at the centre is $\angle AOB = 96^\circ$, find:
(i) The angle subtended by arc $BC$ at the centre ($\angle BOC$).
(ii) The angle subtended at the circumference $\angle CAB$.
(iii) The angle $\angle ADB$ where $D$ lies on the major arc.

(i) Find Central Angle $\angle BOC$: By Arc-Angle Proportionality: $$ \frac{\angle AOB}{\angle BOC} = \frac{\text{arc } AB}{\text{arc } BC} = \frac{3}{2} $$ $$ \frac{96^\circ}{\angle BOC} = \frac{3}{2} \implies \angle BOC = \frac{2 \times 96^\circ}{3} = \mathbf{64^\circ} $$

(ii) Find $\angle CAB$: Arc $BC$ subtends angle $\angle BOC = 64^\circ$ at centre and $\angle CAB$ at the circumference: $$ \angle CAB = \frac{1}{2} \angle BOC = \frac{1}{2} \times 64^\circ = \mathbf{32^\circ} $$

(iii) Find $\angle ADB$: Angle subtended by arc $AB$ at circumference is $\angle ACB = \frac{1}{2}\angle AOB = \frac{96^\circ}{2} = 48^\circ$.
In cyclic quad $ACBD$, opposite angles are supplementary: $$ \angle ADB = 180^\circ - \angle ACB = 180^\circ - 48^\circ = \mathbf{132^\circ} $$

✍ PRACTICE PROBLEM D2: INSCRIBED REGULAR HEXAGON & SUBTENDED ANGLES

Q. A regular hexagon $ABCDEF$ is inscribed in a circle with centre $O$. Calculate:
(i) $\angle AOB$ (Central angle subtended by one side).
(ii) $\angle ACB$ (Angle subtended by side $AB$ at vertex $C$).
(iii) $\angle ADB$ (Angle subtended by side $AB$ at vertex $D$).
(iv) The interior angle $\angle AEF$ of the hexagon.

(i) Central Angle Subtended by Side $AB$: For regular hexagon ($n = 6$): $$ \angle AOB = \frac{360^\circ}{6} = \mathbf{60^\circ} $$

(ii) Angle $\angle ACB$ at Circumference: Side $AB$ subtends $\angle AOB = 60^\circ$ at centre, so at vertex $C$ on the circumference: $$ \angle ACB = \frac{1}{2}\angle AOB = \frac{1}{2} \times 60^\circ = \mathbf{30^\circ} $$

(iii) Angle $\angle ADB$ (Angles in Same Segment): Both $\angle ADB$ and $\angle ACB$ are in the same segment of chord $AB$: $$ \angle ADB = \angle ACB = \mathbf{30^\circ} $$

(iv) Interior Angle $\angle AEF$: Formula for each interior angle of a regular $n$-gon: $$ \text{Interior Angle} = \frac{(n - 2) \times 180^\circ}{n} = \frac{(6 - 2) \times 180^\circ}{6} = \frac{720^\circ}{6} = \mathbf{120^\circ} $$

PART (E): ICSE Board Examination Masterclass (Advanced Multi-Concept Exemplars)

✍ BOARD EXEMPLAR E1: CONCENTRIC CIRCLES & MID-POINT THEOREM PROOF

Q. In a circle with centre $O$, diameter $AB$ and chord $AD$ are drawn. Another circle drawn on $AO$ as diameter cuts $AD$ at $C$. Prove that $BD = 2 \times OC$.

Step 1 (Angle in semi-circle of smaller circle): $AO$ is diameter of the smaller circle, so $\angle ACO = 90^\circ$ (by Theorem 7) $\implies OC \perp AD$.

Step 2 (Chord bisector in larger circle): In the larger circle, line $OC$ from centre $O$ is perpendicular to chord $AD$. By Theorem 2, $OC$ bisects $AD \implies AC = CD \implies C$ is the mid-point of $AD$.

Step 3 (Mid-Point Theorem in $\Delta ABD$): $O$ is the mid-point of diameter $AB$, and $C$ is the mid-point of $AD$.
By Mid-point Theorem in $\Delta ABD$: $$ OC \parallel BD \quad \text{and} \quad OC = \frac{1}{2}BD \implies \mathbf{BD = 2 \times OC} $$ Hence Proved!

✍ BOARD EXEMPLAR E2: CENTRAL ANGLE & CYCLIC EQUATION PROOF

Q. In a circle with centre $O$, $\angle AOC = 160^\circ$. If $y$ is an angle subtended on the major arc and $x$ is an angle subtended on the minor arc by chord $AC$, prove that $3y - 2x = 140^\circ$.

Step 1: Calculate $y$ (Major Arc Angle): By Theorem 5: $$ y = \frac{1}{2}\angle AOC = \frac{1}{2} \times 160^\circ = 80^\circ $$

Step 2: Calculate $x$ (Minor Arc Angle): Opposite angles of cyclic quadrilateral are supplementary: $$ x + y = 180^\circ \implies x = 180^\circ - 80^\circ = 100^\circ $$

Step 3: Evaluate Expression: $$ 3y - 2x = 3(100^\circ) - 2(80^\circ) = 300^\circ - 160^\circ = \mathbf{140^\circ} $$ Hence Proved!

✍ BOARD EXEMPLAR E3: TWO INTERSECTING CIRCLES & CYCLIC PROPERTIES

Q. Two unequal circles with centres $A$ and $B$ intersect at $C$ and $D$. The centre $B$ lies on the bigger circle. If a point $M$ on the smaller circle gives $\angle CMD = x^\circ$, find $\angle DAC$ in terms of $x$.

In Smaller Circle (Centre $B$): Central angle subtended by arc $CD$ is double the circumference angle: $$ \angle CBD = 2 \angle CMD = 2x^\circ $$

In Bigger Circle (Centre $A$): Point $B$ lies on the circumference of the bigger circle. For points $D, B, C, N$ on the bigger circle, $DBCN$ is a cyclic quadrilateral: $$ \angle DNC = 180^\circ - \angle CBD = 180^\circ - 2x^\circ $$

Central Angle of Bigger Circle: $$ \angle DAC = 2 \angle DNC = 2(180^\circ - 2x^\circ) = \mathbf{(360 - 4x)^\circ} $$

✍ BOARD EXEMPLAR E4: TRIANGLE ANGLE BISECTORS & CIRCUMCIRCLE

Q. In $\Delta ABC$, $\angle BAC = 56^\circ$ and $\angle ABC = 64^\circ$. The bisectors of $\angle A, \angle B, \angle C$ meet the circumcircle at $P, Q, R$ respectively. Calculate $\angle QPR$.

Step 1: Find $\angle ACB$: $$ \angle ACB = 180^\circ - (56^\circ + 64^\circ) = 180^\circ - 120^\circ = 60^\circ $$

Step 2: Angles in Same Segments:

$\angle QPA = \angle QBA = \frac{1}{2}\angle ABC = \frac{1}{2}(64^\circ) = 32^\circ$ (subtended by arc $QA$).
$\angle RPA = \angle RCA = \frac{1}{2}\angle ACB = \frac{1}{2}(60^\circ) = 30^\circ$ (subtended by arc $RA$).

Step 3: Calculate $\angle QPR$: $$ \angle QPR = \angle QPA + \angle RPA = 32^\circ + 30^\circ = \mathbf{62^\circ} $$

✍ BOARD EXEMPLAR E5: INCENTRE & CIRCUMCIRCLE INTERSECTIONS

Q. $I$ is the incentre of $\Delta ABC$. $AI$ produced meets the circumcircle of $\Delta ABC$ at $D$. If $\angle BAC = 50^\circ$ and $\angle ABC = 70^\circ$, calculate (i) $\angle BCD$, (ii) $\angle ICD$, (iii) $\angle BIC$.

Third angle $\angle ACB$: $\angle ACB = 180^\circ - (50^\circ + 70^\circ) = 60^\circ$.

(i) Find $\angle BCD$: Line $AD$ bisects $\angle BAC \implies \angle BAD = \frac{50^\circ}{2} = 25^\circ$. $$ \angle BCD = \angle BAD = \mathbf{25^\circ} \quad \text{(angles in same segment of arc } BD\text{)} $$

(ii) Find $\angle ICD$: $CI$ bisects $\angle ACB \implies \angle ICB = \frac{60^\circ}{2} = 30^\circ$. $$ \angle ICD = \angle ICB + \angle BCD = 30^\circ + 25^\circ = \mathbf{55^\circ} $$

(iii) Find $\angle BIC$: In $\Delta IBC$: $$ \angle BIC = 180^\circ - (\angle IBC + \angle ICB) = 180^\circ - (35^\circ + 30^\circ) = \mathbf{115^\circ} $$

✍ BOARD EXEMPLAR E6: PERPENDICULAR CHORDS & ALGEBRAIC EQUALITY

Q. In a circle with centre $O$, chords $AC$ and $BD$ are perpendicular to each other. $\angle OAB = a$ and $\angle DBC = b$. Prove that $a = b$.

Step 1 (In $\Delta AOB$): $OA = OB$ (radii) $\implies \angle OBA = \angle OAB = a$. $$ \angle AOB = 180^\circ - 2a $$

Step 2 (Angle at circumference): $$ \angle ACB = \frac{1}{2} \angle AOB = \frac{180^\circ - 2a}{2} = 90^\circ - a $$

Step 3 (Right angle from perpendicular chords): Chords $AC \perp BD$, so in the right-angled triangle formed at their intersection: $$ \angle ACB + \angle DBC = 90^\circ \implies (90^\circ - a) + b = 90^\circ \implies \mathbf{a = b} $$ Hence Proved!