#### Question

Prove that the diagonals of a rhombus are perpendicular bisectors of each other.

#### Solution

Let OABC be a rhombus, whose diagonals OB and AC intersect at D. Suppose O is the origin.

Let the position vector of A and C be \[\vec{a}\] and \[\vec{c}\] respectively.

Then, \[\vec{OA} = \vec{a}\] In âˆ†OAB, \[\vec{OB} = \vec{OA} + \vec{AB} = \vec{OA} + \vec{OC} = \vec{a} + \vec{c}\]

\[\left( \vec{AB} = \vec{OC} \right)\]

Position vector of mid-point of \[\vec{OB} = \frac{1}{2}\left( \vec{a} + \vec{c} \right)\]

Position vector of mid-point of\[\vec{OB} = \frac{1}{2}\left( \vec{a} + \vec{c} \right)\] (Mid-point formula)

So, the mid-points of OB and AC coincide. Thus, the diagonals OB and AC bisect each other.

Now,

\[\vec{OB} . \vec{AC} = \left( \vec{a} + \vec{c} \right) . \left( \vec{c} - \vec{a} \right)\]

\[ = \left( \vec{c} + \vec{a} \right) . \left( \vec{c} - \vec{a} \right)\]

\[ = \left| \vec{c} \right|^2 - \left| \vec{a} \right|^2 \]

\[ = \left| \vec{OC} \right|^2 - \left| \vec{OA} \right|^2 \]

\[ = 0 \left( \left| \vec{OC} \right| = \left| \vec{OA} \right| \right)\]

\[ \Rightarrow \vec{OB} \perp \vec{AC}\]

Hence, the diagonals OB and AC are perpendicular to each other.

Thus, the diagonals of a rhombus are perpendicular bisectors of each other.