Circle Passing through 3 points, Cyclic Quadrilateral and Related Theorems, Class 9, Mathematics | Extra Documents & Tests for Class 9 PDF Download

Circle through 3 Points

Prove that one and only one circle, passing through three non-collinear points.
 Given :Three non-collinear points A, B, C.
To prove : One and only one circle can be drawn, passing through A, B, and C.

Construction : Join AB and BC. Draw the perpendicular bisectors of AB O and BC. Let these perpendicular bisector intersect meeting at a point O.
Proof :

Hence, one and only circle can be drawn through three non-collinear points A, B and C.

Cyclic Quadrilateral : If all the four vertices of a quadrilateral lie on a circle, then such a quadrilateral is called a cyclic quadrilateral.

If four points lie on a circle, they are said to be concyclic. We also say that quad. ABCD is inscribed in a circle with centre O.

Theorem-8. The opposite angles of a quadrilateral inscribed in a circle are supplementary.

OR

The sum of the opposite angles of a cyclic quadrilateral is 180°.
 Given : A quadrilateral ABCD inscribed in a circle with centre O.
To prove : ∠ADC + ∠ABC = 180° and ∠BAD + ∠BCD = 180°. 

Construction : Join OA and OC.
Proof :

STATEMENT	REASON
1. Arc ABC subtends ∠AOC at the centre and ∠ADC at a point D on the remaining part of the circle. ∠AOC = 2 ∠ADC ⇒∠ADC = 1/2∠AOC	Angle at the centre is double the angle at any point on remaining part of the circle.
	Same as above
	(∠AOC+ reflex ∠AOC)=360°
	= sum of the angles around a point O

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

Converse of above Theorem : If a pair of opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic.
 Given : A quadrilateral ABCD in which ∠B + ∠D = 180°.
To prove : ABCD is a cyclic quadrilateral.
Construction : If possible, let ABCD be not a cyclic
quadrilateral. Draw a circle passing through three

non-collinear points A, B, C. Suppose this circle meets CD or CD produced at D', as shown in Fig. (i) and Fig. (ii) respectively. Join D'A.
Proof :

Hence proved

Theorem-9. The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
 Given :A cyclic quadrilateral whose side AB is produced to a point E.
To prove : ∠CBE = ∠ADC

Proof :

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

Ex. In the given circle with diameter AB, find the value of x.

Sol. 

Ex. If O is the centre of the circle, find the value of x in each of the following figures, giving reasons.

Sol. We have:

(i) ∠BDC = ∠BAC = 35° [∠s in the same segment].

In ΔDBC, we have

∠DBC + ∠BCD + ∠BDC = 180° [Sum of the ∠s of a Δ is 180°]

⇒ 70° + x° + 35° = 180°

⇒ x° = (180° – 105°) = 75°.

Hence, x = 75°.

(ii) ∠BCD = 90°. [Angle in a semi-circle]

In ΔBCD, we have

∠BCD + ∠CDB + ∠DBC = 180° [Sum of the ∠s of a Δ is 180°]

⇒ 90° + ∠CDB + 55° = 180°

⇒ ∠CDB = (180° – 145°) = 35°.

∴ x = ∠BAC = ∠CDB = 35°. [∠s in the same segment]

Hence, x = 35°.

(iii) Reflex ∠AOC = (360 – 110)° = 250°.

Major arc CA subtends reflex ⇒AOC at the centre and ⇒ABC at a point B on the remaining part of the circle.

Hence, x = 125°.

(iv) Since ABD is a straight line, we have

∠ABC + ∠CBD = 180°

⇒ ∠ABC + 70° = 180°

⇒ ∠ABC = (180° – 70°) = 110°.

Reflex ∠AOC = 2 ∠ABC = (2 × 110°) = 220°.

[Major arc CA subtends reflex ∠AOC at the centre and ABC at a point B on remaining part of the circle.]

∴ x = 220°.

Ex.20 In the figure, AB is a diameter of a circle with centre O and CD ║ BA.

If ∠BAC = 20°, find the values of
 (i) ∠BOC 

(ii) ∠DOC 

(iii) ∠DAC 

(iv) ∠ADC.

Sol. (i) Arc BC subtends ∠BOC at the centre and ∠BAC at the circumference.

∴ ∠BOC = 2∠BAC = (2 × 20°) = 40°. [∵ Angle at the centre is double the angle at circumference]

(ii) ∠OCD = ∠BOC = 40° [Alt. interior ∠s as CD ║ BA]

Now, OC = OD [Radii of the same circle]

⇒ ∠ODC = ∠OCD = 40°.

In ∠OCD, we have

∠DOC + ∠OCD + ∠ODC = 180° [Sum of the ∠s of a Δ is 180°]

⇒ ∠DOC + 40° + 40° = 180°

⇒ ∠DOC = (180° – 80°) = 100°.

(iii) ∠DAC

=     = 50°.        [Angle at the centre is double the ∠ at circumference]

(iv) ∠ACD = ∠CAB = 20° [Alt. Int. ∠s, as CD ║ BA]

In ΔACD, we have

∠ADC + ∠ACD + ∠DAC = 180° [Sum of the ∠s of a Δ is 180°]

⇒ ∠ADC + 20° + 50° = 180°

⇒ ∠ADC = (180° – 70°) = 110°.

Hence, ∠BOC = 40°, ∠DOC = 100°, ∠DAC = 50° and ∠ACD = 110°.