Given: AB and CD are two chords of a circle with centre O, intersecting at point E. PQ is a diameter through E, such that ∠AEQ = ∠DEQ. To prove: AB = CD Construction: Draw OL perpendicular AB and OM perpendicular CD Proof: ∠LOE + ∠LEO + ∠OLE = 180° (Angle sum property of a triangle) ⇒ ∠LOE + ∠LEO = 90° ...(i) Similarly ∠MOE + ∠MEO + ∠OME = 180° ⇒ ∠MOE + ∠MEO + 90° = 180° ∠MOE + ∠MEO = 90° ...(ii) From (i) and (ii) we get ∠LOE + ∠LEO = ∠MOE + ∠MEO ...(iii) Also, ∠LEO = ∠MEO (Given) ...(iv) From (i) and (iv) we get ∠LOE = ∠MOE Now in triangles OLE and OME ∠LEO = ∠MEO (Given) ∴ ∠LOE = ∠MOE (Proved above) EO = EO (Common) ∴ △OLE ≅ △OME (ASA congruence criterion) ∴ OL = OM (CPCT) Thus, chords AB and CD are equidistant from the centre of the circle. Since, chords of a circle which are equidistant from the centre are equal. ∴ AB = CD