# Show by using vectors that the two diagonals of a square are equal in dimension

and also perpendicular to each other?

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Let |A| be Euclidean length of vector A

Now as per Pythagoras theorem: For two orthogonal vector - say A & B - in 2D,
|A+B| = sqrt(|A|^2 + |B|^2)

Now, let a, b, c, and d are the corner point of given square in clockwise direction and ab, bc, cd, and da are vectors which represents the sides of square.
So, |ab| = |bc| = |cd| = |da|

So, dimension of diagonal:
{Dimension_of ac or ca} = |ac| = sqrt(|ab|^2 + |bc|^2) = sqrt(|ab|^2 + |ad|^2) = |bd| = {Dimension_of bd or db}

Above argument is correct for 2D. and Square is a planar figure, so we can apply these rules.

So, this proves that, two diagonals of a square are equal in dimension.
Done!

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A hint. We have a general arrangement like this

If we let $\vec{BA} = \begin{pmatrix} a \\ b \end{pmatrix}$ then what will the vector $\vec{DA}$ be? How do we represent a vector perpendicular to a given vector?

What about $\vec{CA}$? And $\vec{BD}$? Can they be written in terms of $\vec{BA}$ and $\vec{DA}$?

How do we determine when two vectors are perpendicular?

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Two vectors are orthogonal if and only if their dot product is 0. Can you use that fact in your proof?

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