Question

Prove that the diagonals of a rhombus bisect each other at right angles.

Answer 1

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.