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

Solution:

Rhombus is a parallelogram.
 

Consider:

$\Delta A O B$ and $\Delta C O D$

$\angle O A B=\angle O C D \quad$ (alternate angle)

$\angle O D C=\angle O B A \quad$ (alternate angle)

$\angle D O C=\angle A O B \quad$ (vertically opposite angles)

$\Delta A O B \cong C O B$

$\therefore A O=C O$

$O B=O D$

Therefore, the diagonals bisects at O.

Now, let us prove that the diagonals intersect each other at right angles.

Consider $\Delta C O D$ and $\Delta C O B$ :

$C D=C B \quad$ (all sides of a rhombus are equal)

$C O=C O \quad$ (common side)

$O D=O B \quad(p$ oint $O$ bisects $B D)$

$\therefore \triangle C O D \cong \Delta C O B$

$\therefore \angle C O D=\angle C O B$        (corresponding parts of congruent triangles)'

Further, $\angle C O D+\angle C O B=180^{\circ} \quad(l$ inear pair $)$

$\therefore \angle C O D=\angle C O B=90^{\circ}$

It is proved that the diagonals of a rhombus are perpendicular bisectors of each other.