Show that the diagonals of a rhombus are perpendicular to each other.
(N/A) Consider the rhombus $ABCD$ (see Fig.).
You know that $AB = BC = CD = DA$ (by definition of a rhombus).
Now,in $\Delta AOD$ and $\Delta COD$,
$OA = OC$ (Diagonals of a parallelogram bisect each other).
$OD = OD$ (Common side).
$AD = CD$ (Sides of a rhombus are equal).
Therefore,$\Delta AOD \cong \Delta COD$ ($SSS$ congruence rule).
This gives,$\angle AOD = \angle COD$ $(CPCT)$.
But,$\angle AOD + \angle COD = 180^{\circ}$ (Linear pair).
So,$2 \angle AOD = 180^{\circ}$.
Or,$\angle AOD = 90^{\circ}$.
Thus,the diagonals of a rhombus are perpendicular to each other.
You know that $AB = BC = CD = DA$ (by definition of a rhombus).
Now,in $\Delta AOD$ and $\Delta COD$,
$OA = OC$ (Diagonals of a parallelogram bisect each other).
$OD = OD$ (Common side).
$AD = CD$ (Sides of a rhombus are equal).
Therefore,$\Delta AOD \cong \Delta COD$ ($SSS$ congruence rule).
This gives,$\angle AOD = \angle COD$ $(CPCT)$.
But,$\angle AOD + \angle COD = 180^{\circ}$ (Linear pair).
So,$2 \angle AOD = 180^{\circ}$.
Or,$\angle AOD = 90^{\circ}$.
Thus,the diagonals of a rhombus are perpendicular to each other.
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