Let O be the center of circle, with AB and CD as equal chords.
From figure,
X is the point of intersection from the chords.
Draw OM ⊥ AB and ON ⊥ CD.
In ∆ OMX and ∆ ONX,
⇒ ∠OMX = ∠ONX = 90°
⇒ OX = OX (Common)
We know that,
AB and CD are equal chords and equal chords are equidistant from the centre.
⇒ OM = ON
⇒ ∆ OMX ≅ ∆ ONX (By R.H.S. congruence rule)
⇒ ∠OXM = ∠OXN (By C.P.C.T.)
⇒ ∠OXA = ∠OXD
Hence, proved that the line joining the point of intersection to the centre makes equal angles with the chords.