if two intersection chords of circle make equal angle with diameter passing thorough their point of intersection. prove that chords 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)