Given : AB and CD are two equal chords of a circle which meet at E within the circle and a line PQ joining the point of intersection to the centre.
To Prove : ∠AEQ = ∠DEQ
Construction : Draw OL ⊥ AB and OM ⊥ CD.
Proof : In ∆OLE and ∆OME, we have
OL = OM [Equal chords are equidistant]
OE = OE [Common]
∠OLE = ∠OME [Each = 90°]
∴ ∆OLE ≅ ∆OME [RHS congruence]
⇒ ∠LEO = ∠MEO [CPCT]