Revision: Geometry >> Angle and Cyclic Properties of a Circle Maths (English Medium) ICSE Class 10 CISCE

A circle is defined as the figure (closed curve) obtained by joining all those points in a plane which are at a constant distance from a fixed point in the same plane. 

• Centre → Fixed point

• Radius → Constant distance

• Circumference → Perimeter of the circle

The line segment, joining any two points on the circumference of the circle, is called a chord. 

Let ABCD be a parallelogram inscribed in a circle.

Now, ∠BAD + ∠BCD

(Opposite angles of a parallelogram are equal.)

And ∠BAD + ∠BCD = 180°

(A pair of opposite angles in a cyclic quadrilateral are supplementary.)

∠BAD + ∠BCD = `(180^circ)/2` = 90°

The other two angles are 90°, and the opposite pair of sides are equal.

∴ ABCD is a rectangle.

Statement:
The angle in a semicircle is a right angle.

Result:

∠ACB = 90^∘

Short Proof (Idea):

• Diameter subtends an angle of 180° at the centre.

• The angle at the circle is half of it.

• Therefore, angle = 90°.

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

Result:

∠ACB = ∠ADB

Short Proof (Idea):

• Both angles stand on the same arc.

• The angle at the centre is double each of them.

• Hence, both angles are equal.

Statement:
The angle subtended by an arc at the centre of a circle is double the angle subtended by it at any point on the remaining part of the circle.

Result:

∠AOB = 2∠ACB

Short Proof (Idea):

• Join the centre to the points on the circle.

• Radii form isosceles triangles.

• Angle at centre = sum of angles at the circle.

• Hence, the angle at the centre is double the angle at the circle.

Statement:
If an arc of a circle subtends a right angle at any point on the remaining part of the circle, then the arc is a semicircle.

Result:
Arc AB is a semicircle
(or AB is a diameter)

Short Proof (Idea):

• Given angle at the circle ∠ACB = 90°.

• The angle at the centre is double the angle at the circle.

• Therefore, ∠AOB = 2 × 90° = 180°.

• Hence, A, O, and B lie on a straight line, so AB is a diameter.

• Therefore, arc AB is a semicircle.

Let two circles having their centres as O and O’ intersect each other at point A and B respectively. Let us join OO’.

In ΔAOO’ and BOO’,

OA = OB   ...(Radius of circle 1)

O’A = O’B   ...(Radius of circle 2)

OO’ = OO’   ...(Common)

ΔAOO’ ≅ ΔBOO’   ...(By SSS congruence rule)

∠OAO’ = ∠OBO’   ...(By CPCT)

Therefore, line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Statement:
The sum of the opposite angles of a cyclic quadrilateral is 180°.

Short Proof (Idea):

• Let ABCD be a cyclic quadrilateral.

• Arc ABC subtends an angle ∠ADC at the circle and ∠AOC at the centre.

• The angle at the centre is double the angle at the circle.

  ∠ADC=`1/2`∠AOC

• Similarly, the other arc subtends:

  ∠ABC = `1/2`(reflex ∠AOC)

• The sum of angles around the centre is 360°.

• Therefore,

  ∠ADC + ∠ABC = `1/2`(360°) = 180^∘

Conclusion:
Hence, the opposite angles of a cyclic quadrilateral are supplementary.

Statement:
If the sum of a pair of opposite angles of a quadrilateral is 180°, then the quadrilateral is cyclic.

Short Proof (Idea):

• Given, ∠B + ∠D = 180°.

• Draw a circle through three vertices of the quadrilateral.

• If the fourth vertex does not lie on the circle, an exterior angle becomes equal to its interior opposite angle, which is not possible.

• Hence, the fourth vertex must lie on the same circle.

Conclusion:
Therefore, ABCD is a cyclic quadrilateral.

Statement:

The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

Short Proof (Idea):

• In a cyclic quadrilateral, the sum of opposite angles is 180°.

• The exterior angle and the adjacent interior angle form a straight line, so their sum is 180°.

• Since both are supplementary to the same angle, they are equal.

Conclusion:

Therefore, the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

In ΔABC,

∠ABC + ∠BCA + ∠CAB = 180°  ...(Angle sum property of a triangle)

⇒ 90° + ∠BCA + ∠CAB = 180°

⇒ ∠BCA + ∠CAB = 90°    ...(1)

In ΔADC,

∠CDA + ∠ACD + ∠DAC = 180°   ...(Angle sum property of a triangle)

⇒ 90° + ∠ACD + ∠DAC = 180°

⇒ ∠ACD + ∠DAC = 90°   ...(2)

Adding equations (1) and (2), we obtain

∠BCA + ∠CAB + ∠ACD + ∠DAC = 180°

⇒ (∠BCA + ∠ACD) + (∠CAB + ∠DAC) = 180°

∠BCD + ∠DAB = 180°     ...(3)

However, it is given that

∠B + ∠D = 90° + 90° = 180°    ...(4)

From equations (3) and (4), it can be observed that the sum of the measures of opposite angles of quadrilateral ABCD is 180°. Therefore, it is a cyclic quadrilateral.

Consider chord CD.

∠CAD = ∠CBD   ...(Angles in the same segment)

Given: `square`ABCD is a cyclic quadrilateral.

To prove: ∠BAD + ∠BCD = 180º 

∠ABC + ∠ADC = 180º

Proof: Arc BCD is intercepted by the inscribed ∠BAD.

∠BAD = `1/2` m(arc BCD)     ...(i) [Inscribed angle theorem]

Arc BAD is intercepted by the inscribed ∠BCD.

∴ ∠BCD = `1/2` m(arc DAB)      ...(ii) [Inscribed angle theorem]

From (1) and (2) we get

∠BAD + ∠BCD = `1/2` [m(arc BCD) + m(arc DAB)]

∴ (∠BAD + ∠BCD) = `1/2 xx 360^circ`    ...[Completed circle]

= 180°

Again, as the sum of the measures of angles of a quadrilateral is 360°

∴ ∠ADC + ∠ABC = 360° – [∠BAD + ∠BCD]

= 360° – 180°

= 180°

Hence, the opposite angles of a cyclic quadrilateral are supplementary.

It is given that BE is the bisector of ∠B.

∴ ∠ABE = ∠B/2

However, ∠ADE = ∠ABE (Angles in the same segment for chord AE)

⇒ ∠ADE = ∠B/2

Similarly, ∠ACF = ∠ADF = ∠C/2         (Angle in the same segment for chord AF)

∠D = ∠ADE + ∠ADF

`= (angleB)/2 + (angleC)/2`

`= 1/2(angleB + angleC)`

`= 1/2(180^@ - angleA)`

`= 90^@ - 1/2angleA`

Similarly, it can be proved that

`angleE = 90^@ - 1/2angleB`

`angleF = 90^@ - 1/2angleC`

Consider a ΔABC.

Two circles are drawn while taking AB and AC as the diameter.

Let them intersect each other at D, and let D not lie on BC.

Join AD.

∠ADB = 90°...(Angle subtended by semi-circle)

∠ADC = 90°   ...(Angle subtended by semi-circle)

∠BDC = ∠ADB + ∠ADC = 90° + 90° = 180°

Therefore, BDC is a straight line, and hence, our assumption was wrong.

Thus, point D lies on the third side BC of ΔABC.

• Concentric circles → Same centre, different radii

• Equal circles → Same radius

• Circumscribed circle → Circle passes through all vertices of a polygon
  Centre → Circumcentre

• Inscribed circle → Circle touches all sides of a polygon
  Centre → Incentre

• Diameter → Longest chord of a circle

• Perpendicular from centre to a chord → Bisects the chord

• Line joining centre to midpoint of a chord → Perpendicular to the chord

• The greater the chord → Nearer to the centre

• The smaller the chord, → farther from the centre

• Equal chords → Equidistant from the centre

• Chords equidistant from centre → Equal in length

• Only one circle passes through three non-collinear points

1. Arc Definition: An arc is a curved portion of a circle's circumference between two points.

2. Two Types: Minor arc (< 180°) and Major arc (> 180°).

3. Semicircle: When the arc angle is exactly 180°, it's called a semicircle.

4. Complete Circle: Minor arc + Major arc = 360° (complete circumference).

Definition: A segment is a region of a circle bounded by a chord and its arc

Two Main Types:

• Minor Segment = smaller piece

• Major Segment = larger piece

Semicircle Special Case:

• Formed when chord = diameter

• Creates two perfectly equal segments

• Each semicircle = half the circle's area

• Intersecting Circles (Collinearity):
  If two circles with centres O and O′ intersect at two points and diameters are drawn through these points, then the given points lie on a straight line.

• Cyclic Parallelogram:
  Every parallelogram inscribed in a circle is a rectangle.

• Isosceles Trapezium:
  An isosceles trapezium is always a cyclic quadrilateral.

• Cyclic Trapezium Properties:
  A cyclic trapezium is isosceles, and its diagonals are equal.

• Regular Polygon:
  Any four vertices of a regular polygon lie on a circle.

• Chords Bisecting Each Other:
  If two chords of a circle bisect each other, then both the chords are diameters of the circle.

• Angle Bisectors of a Cyclic Quadrilateral:
  The quadrilateral formed by the angle bisectors of a cyclic quadrilateral is also cyclic.

• Mid-point of Hypotenuse:
  The mid-point of the hypotenuse of a right-angled triangle is equidistant from all three vertices of the triangle.

• Equal Chords and Angles at the Centre:
  In a circle, equal chords subtend equal angles at the centre.

• Converse of Equal Chords Theorem:
  In a circle, chords which subtend equal angles at the centre are equal in length.