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Question
Prove that: If the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Solution
Let OACB be a quadrilateral such that diagonals OC and AB bisect each other at 90º.
Taking O as the origin, let the poisition vectors of A and B be \[\vec{a}\] and \[\vec{b}\] respectively.
Then, \[\vec{OA} = \vec{a}\] and \[\vec{OB} = \vec{b}\] Position vector of mid-point of AB, \[\vec{OE} = \frac{\vec{a} + \vec{b}}{2}\]
∴ Position vector of C, \[\vec{OC} = \vec{a} + \vec{b}\]
By the triangle law of vector addition, we have
\[\vec{OA} + \vec{AB} = \vec{OB} \]
\[ \Rightarrow \vec{AB} = \vec{OB} - \vec{OA} = \vec{b} - \vec{a}\]
Since \[\vec{AB} \perp \vec{OC}\]
\[\Rightarrow \vec{AB} . \vec{OC} = 0\]
\[ \Rightarrow \left( \vec{b} - \vec{a} \right) . \left( \vec{a} + \vec{b} \right) = 0\]
\[ \Rightarrow \left| \vec{b} \right|^2 - \left| \vec{a} \right|^2 = 0\]
\[ \Rightarrow \left| \vec{a} \right|^2 = \left| \vec{b} \right|^2 \]
\[ \Rightarrow \left| \vec{a} \right| = \left| \vec{b} \right|\]
\[ \Rightarrow OA = OB\]
In a quadrilateral if diagonals bisects each other at right angle and adjacent sides are equal, then it is a rhombus.
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