Circle Theorems and ProofsTheorem 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 = ∠POQHence, 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 ODProof: Given, In ∆OPB and ∆OQDBP = 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 = CDTheorem 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 4From the above figure,∠AOB = 2∠APBConstruction: Join PD passing through centre OProof: 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 + ∠BODFrom eq. 4 and 5, we get;⇒ ∠AOB = 2∠OPA + 2∠OPB⇒ ∠AOB = 2(∠OPA + ∠OPB)⇒ ∠AOB = 2∠APBHence, 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:hlo.. good morning ☺️hru??