Solution:
Given two equal chords of a circle.Let us assume AB and AC are two equal chords of a circle with center O and, AD intersect BC at point P.
AB=AC
BAP CAP
AP=BP (common side)
By SAS congruence rule, BAP CAP
And by CPCT theorem,
BAP CPA
Since perpendicular bisector AP bisect chord BC, then CP=CB.
As BPA and CPA are linear pair angles, so BPA+ CPA , then BPA CPA .
Thus, the perpendicular bisector AP of chord BC is passing through the center O.
Hence, the statement is true and option 1 is correct.