Let ABCD be a rhombus whose diagonal AC and BD intersect at the point O.
As we know that the diagonals of a parallelogram bisect each other and rhombus is a parallelogram.
So, OA = OC and OB = OD.
From ∆ COB and ∆ COD we get,
CB = CD (sides of rhombus) and
CO is common in both the triangles.
So, OB = OD
Therefore, by SSS theorem.
∆ COB ≅ ∆ COD
\(\angle COB\) = \(\angle COD\)
\(\angle COB\) + \(\angle COD\) = 180° (Linear pair of angles)
Thus, \(\angle COB\) = \(\angle COD\) = 90°
Hence, the diagonals of a rhombus bisect each other at right angles.