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If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords

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 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]

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