To prove that the perpendicular from the centre to a chord bisect the chord.Consider a circle with centre at \\\\(O\\\\) and \\\\(AB\\\\) is a chord such that \\\\(OX\\\\) perpendicular to \\\\(AB\\\\)To prove that \\\\(AX = BX\\\\)In \\\\( \\Delta OAX \\\\) and \\\\( \\Delta OBX \\\\)\\\\( \\angle OXA = \\angle OXB \\\\) [both are 90 ]\\\\( OA = OB \\\\) (Both are radius of circle )\\\\( OX = OX \\\\) (common side )\\\\(ΔOAX\\cong ΔOBX\\\\)\\\\(AX = BX\\\\) (by property of congruent triangles )hence proved.