Theorems on Chord
In this section we will discuss Theorems on Chord.
Theorems on Circles at both the Intermediate and Higher tier are difficult areas for students, which teachers often find equally difficult to deliver.Here, I have explained some Circle theorems based on circle so that student can understand them easily.
1) If two arcs of a circle are congruent, then corresponding chords are equal.
Given : PQ = BC
Prove that : ∠POQ = ∠BAC
1) The perpendicular from the center of a circle to a chord bisects the chord.
Given : PQ is a chord of a circle and OL ⊥ PQ.
Prove that : LP = LQ
2) If two chords of a circle AB and AC of a circle with center O are such that center O lies on the bisector of ∠BAC, then AB = AC ( chords are equal ).
Given : AB = AC
Prove that : Center O lies on the bisector of ∠BAC
Theorems on Circle
• Theorems on Chord
• Theorems on Chord and Subtended Angle
• Theorems on Arc and Angle
• Theorems on Cyclic Quadrilateral
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Theorems on Circles at both the Intermediate and Higher tier are difficult areas for students, which teachers often find equally difficult to deliver.Here, I have explained some Circle theorems based on circle so that student can understand them easily.
1) If two arcs of a circle are congruent, then corresponding chords are equal.
Given : PQ = BC
Prove that : ∠POQ = ∠BAC



1) BC = PQ  1) Given  
2) OP = AB  2) Radius of a circle  
3) OQ = AC  3) Radius of a circle  
4) ΔPOQ = ΔBAC  4) SSS Postulate  
5) ∠POQ = ∠BAC  5) CPCTC  
Given : PQ is a chord of a circle and OL ⊥ PQ.
Prove that : LP = LQ



1) OP = OQ  1) Radii of the same circle.  
2) OM = OM  2) Reflexive (common)  
3) ∠OMP = ∠OMQ  3) Each 90^{0}  
4) ΔPMO = ΔQMO  4) HL postulate(RHS)  
5) PM = MQ  5) CPCTC  
Given : AB = AC
Prove that : Center O lies on the bisector of ∠BAC



1) AB = AC  1) Given  
2) ∠BAM = ∠CAM  2) Given  
3) AM = AM  3) Reflexive (Common)  
4) ΔBAM = ΔCAM  4) SAS Postulate  
5) BM = CM and ∠BMA = ∠CMA  5) CPCTC  
6) ∠BMA + ∠CMA = 90^{0}  6) Linear pair angles  
7) AM = BM and ∠BMA = ∠CMA = 90^{0}  7) From (6)  
8) AM is the perpendicular bisector of BC  8) Definition of perpendicular bisector.  
9) AM passes through the center O.  9) Perpendicular bisector of a chord always passes through the center.  
Theorems on Circle
• Theorems on Chord
• Theorems on Chord and Subtended Angle
• Theorems on Arc and Angle
• Theorems on Cyclic Quadrilateral