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.
1. O lies on the perpendicular bisector of AB
Every point on perpendicular bisector of a line segment
is equidistant from its end points i.e. A and B.
2. O lies on the perpendicular bisector of B
⇒ OB = OC... (ii)
Each point on perpendicular bisector of line segment is
equidistant from its end points i.e. B and C.
3. OA = OB = OC
⇒ O is equidistant from A, B and C
|From (i) and (ii)|
|4. O is the only point equidistant from A, B andC.||Perpendicular bisectors of AB and BC cut each other at|
point O only
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.
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.
1. Arc ABC subtends ∠AOC at the centre and ∠ADC at a point D on the remaining part of the circle.
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.
1. ∠B + ∠D = 1800
|2. ∠B + ∠D' = 1800||ABCD' is a cyclic quadrilateral and so its opposite ∠s are supplementary.|
|3. ∠B + ∠D = ∠B + ∠D'|
= ∠D = ∠D'
|From (i) and (ii).|
|4. But, this is not possible. A|
∴Our supposition is wrong.
ABCD is a cyclic quadrilateral.
|An exterior angle of a Δ is never equal to its into opp. angle.|
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
1. ∠ABC + ∠ADC = 1800
2. ∠ABC + ∠CBE = 1800
3. ∠ABC + ∠ADC = ∠ABC + ∠CBE
ABCD is a cyclic quadrilateral and so the sum of its opp.∠s is 180°.
ABE is a straight line.
From (i) and (ii)
The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
Ex. In the given circle with diameter AB, find the value of x.
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
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°.
= = 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°.
|1. What is a cyclic quadrilateral?|
|2. How many points are required to uniquely determine a circle passing through them?|
|3. How can we prove that a quadrilateral is cyclic?|
|4. Can a triangle be a cyclic quadrilateral?|
|5. What are some properties of a cyclic quadrilateral?|