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Question
Prove that the diagonals of a rectangle are perpendicular if and only if the rectangle is a square.
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Solution
Let ABCD be a rectangle. Take A as the origin.
Suppose the position vectors of points B and D be \[\vec{a}\] and \[\vec{b}\] respectively.
Now,
\[\vec{AC} = \vec{AB} + \vec{BC} = \vec{AB} + \vec{AD} = \vec{a} + \vec{b}\]
Also,
\[\vec{BD} = \vec{a} - \vec{b}\]
Since ABCD is rectangle, so \[\vec{AB} \perp \vec{AD}\]
\[\therefore \vec{a} . \vec{b} = 0\]
Now, diagonals AC and BD are perpendicular iff \[\vec{AC} . \vec{BD} = 0\]
\[\text{ iff } \left( \vec{a} + \vec{b} \right) . \left( \vec{a} - \vec{b} \right) = 0\]
\[\text{ iff } \left| \vec{a} \right|^2 - \left| \vec{b} \right|^2 = 0\]
\[\text{ iff }\left| \vec{a} \right| = \left| \vec{b} \right|\]
\[\text{ iff } \left| \vec{AB} \right| = \left| \vec{AD} \right|\]
iff ABCD is a square
Thus, the diagonals of a rectangle are perpendicular if and only if the rectangle is a square.
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