if two intersection chords of circle make equal angle with diameter passing thorough their point of intersection. prove that chords are equal
Asked by Sanjay kumar Shukla | 3rd Dec, 2013, 09:02: PM
Given: AB is the diameter of the circle with centre O. AP and AQ are two intersecting chords of the circle such that ∠PAB = ∠QAB.
To prove: AP = AQ
Construction: Draw OL⊥AB and OM⊥AC.
Proof : In ?AOL and ?AOM
∠OLA = ∠OMB (each 90°)
OA = OA (Common line)
∠OAL = ∠OAM (∠PAB = ∠QAB)
Therefore, ?AOL ≅ ?AOM (AAS congruence criterion)
OL = OM (C.P.C.T)
Chords AP and AQ are equidistant from centre O
AP = AQ (Chords which are equidistant from the centre are equal)

Given: AB is the diameter of the circle with centre O. AP and AQ are two intersecting chords of the circle such that ∠PAB = ∠QAB.
To prove: AP = AQ
Construction: Draw OL⊥AB and OM⊥AC.
Proof : In ?AOL and ?AOM
∠OLA = ∠OMB (each 90°)
OA = OA (Common line)
∠OAL = ∠OAM (∠PAB = ∠QAB)
Therefore, ?AOL ≅ ?AOM (AAS congruence criterion)
OL = OM (C.P.C.T)
Chords AP and AQ are equidistant from centre O
AP = AQ (Chords which are equidistant from the centre are equal)
Answered by | 3rd Dec, 2013, 10:42: PM
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Sign Up
Verify mobile number
Enter the OTP sent to your number
Change