Circle through 3 Points
Prove that one and only one circle, passing through three noncollinear points.
Given :Three noncollinear 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 :
STATEMENT  REASON 
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 noncollinear 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.
Theorem8. 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.  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
noncollinear 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 :
STATEMENT  REASON 
1. ∠B + ∠D = 180^{0}  Given. 
2. ∠B + ∠D' = 180^{0}  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. 
Hence proved
Theorem9. 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 :
STATEMENT  REASON 
1. ∠ABC + ∠ADC = 180^{0} 2. ∠ABC + ∠CBE = 180^{0} 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.
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 semicircle]
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°.
1 videos228 docs21 tests

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? 
1 videos228 docs21 tests


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