all the thorems you have learnt in class 9 of mathematics
Answers
Answer:
Step-by-step explanation:
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°
Answer:
“The perpendicular to a chord bisects the chord if drawn from the center 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°