explain the all the theorem of circle.
Answers
Answer:
Circle Theorems and Proofs
Theorem 1:
“Two equal chords of a circle subtend equal angles at the centre of the circle.
Circle Theorem 1
Proof: Given, in ∆AOB and ∆POQ,
AB = PQ (Equal Chords) ______(1)
OA = OB= OP=OQ (Radii of the circle)_____(2)
From eq 1 and 2, we get;
∆AOB ≅ ∆POQ (SSS Axiom of congruency)
Therefore, by CPCT (corresponding parts of congruent triangles), we get;
∠AOB = ∠POQ
Hence, Proved.
Converse of Theorem 1:
“If two angles subtended at the center by two chords are equal, then the chords are of equal length.”
Proof: Given, in ∆AOB and ∆POQ,
∠AOB = ∠POQ (Equal angle subtended at centre O) _______(1)
OA = OB = OP = OQ (Radii of the same circle) ______(2)
From eq. 1 and 2, we get;
∆AOB ≅ ∆POQ (SAS Axiom of congruency)
Hence,
AB = PQ (By CPCT)
Theorem 2:
“The perpendicular to a chord bisects the chord if drawn from the centre of the circle.”
circle theorem 2
In the above figure, as per the theorem, OD ⊥ AB, therefore, AD = DB.
Proof: Given, in ∆AOD and ∆BOD,
∠ADO = ∠BDO = 90° (OD ⊥ AB)________(1)
OA = OB (Radii of the circle) _______(2)
OD = OD (Common side) ______(3)
From eq. (1), (2) and (3), we get;
∆AOB ≅ ∆POQ (R.H.S Axiom of congruency)
Hence, AD = DB (By CPCT)
Converse of Theorem 2:
“A straight line passing through the centre of a circle to bisect a chord is perpendicular to the chord.”
Proof: Given, in ∆AOD and ∆BOD,
AD = DB (OD bisects AB) ________(1)
OA = OB (Radii of the circle) _____(2)
OD = OD (Common side) ________(3)
From eq. 1, 2 and 3, we get;
∆AOB ≅ ∆POQ (By SSS Axiom of congruency)
Hence,
∠ADO = ∠BDO = 90° (By CPCT)
Theorem 3:
“Equal chords of a circle are equidistant (equal distance) from the centre of the circle.”
Circle Theorem 3
Construction: Join OB and OD
Proof: Given, In ∆OPB and ∆OQD
BP = 1/2 AB (Perpendicular to a chord bisects it______(1)
DQ = 1/2 CD (Perpendicular to a chord bisects it____(2)
AB = CD (Given)
BP = DQ (from eq 1 and 2)
OB = OD (Radii of the same circle)
∠OPB = ∠OQD = 90° (OP ⊥ AB and OQ ⊥ CD)
∆OPB ≅ ∆OQD ( By R.H.S Axiom of Congruency)
Hence,
OP = OQ ( By CPCT)
Converse of Theorem 3:
“Chords of a circle, which are at equal distances from the centre are equal in length, is also true.”
Proof: Given, in ∆OPB and ∆OQD,
OP = OQ ________(1)
∠OPB = ∠OQD = 90° ______.(2)
OB = OD (Radii of the same circle) _____(3)
Therefore, from eq. 1, 2 and 3, we get;
∆OPB ≅ ∆OQD (By R.H.S Axiom of Congruency)
BP = DQ ( By CPCT)
1/2 AB = 1/2 CD (Perpendicular from center bisects the chord)
Hence,
AB = CD
Theorem 4:
“Measure of angles subtended to any point on the circumference of the circle from the same arc is equal to half of the angle subtended at the center by the same arc.”
Circle Theorem 4
From the above figure,
∠AOB = 2∠APB
Construction: Join PD passing through centre O
Proof: In ∆AOP,
OA = OP (Radii of the same circle) _____(1)
∠OAP = ∠OPA (Angles opposite to equal sides of a triangle) ______(2)
∠AOD = ∠OAP + ∠OPA (Exterior Angle Property of the triangle) ______(3)
Hence, from eq. 2 and 3 we get;
∠AOD = 2∠OPA______(4)
Similarly in ∆BOP,
Exterior angle, ∠BOD = 2 ∠OPB ______(5)
∠AOB = ∠AOD + ∠BOD
From eq. 4 and 5, we get;
⇒ ∠AOB = 2∠OPA + 2∠OPB
⇒ ∠AOB = 2(∠OPA + ∠OPB)
⇒ ∠AOB = 2∠APB
Hence, proved.
Theorem 5:
“The opposite angles in a cyclic quadrilateral are supplementary.”
Circle Theorem 5
Proof:
Suppose, for arc ABC,
∠AOC = 2∠ABC = 2α (Theorem 4) ______(1)
Consider for arc ADC,
Reflex ∠AOC = 2 ∠ADC = 2β (Theorem 4) ______.(2)
∠AOC + Reflex ∠AOC = 360°
From eq. 1 and 2, we get;
⇒ 2 ∠ABC + 2∠ADC = 360°
⇒ 2α + 2β = 360°
⇒α + β = 180°
Step-by-step explanation:
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Answer:
The perpendicular from the centre of a circle to a chord will always bisect the chord (split it into two equal lengths). Angles formed from two points on the circumference are equal to other angles, in the same arc, formed from those two points.
Circle Theorem 1 - Angle at the Centre.
Circle Theorem 2 - Angles in a Semicircle.
Circle Theorem 3 - Angles in the Same Segment.
Circle Theorem 4 - Cyclic Quadrilateral.
Circle Theorem 5 - Radius to a Tangent.
Circle Theorem 6 - Tangents from a Point to a Circle.
Circle Theorem 7 - Tangents from a Point to a Circle II.