Let us draw two circles of same radius which are passing through the centres of the other circle.
Here, point A and B are the centres Of these circles and these circles are intersecting
each other at point c and O.
In quadrilateral ADBC,
AD = AC (Radius of circle centered at A)
BC = BD (Radius of circle centered at B)
As radius of both circles are equal, therefore, AD = AC = BC = BD
Hence, ADBC is a rhombus and i an rhombus, the diagonals bisect each other at \\( 90 ^ { \\circ } \\).
Hence, \\( \\overline { \\mathrm { AB } } \\) and \\( \\overline { \\mathrm { CD } } \\) are at right angles.