** 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]