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You could prove this by congruent triangles, but here are two simpler arguments:

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Since a square is a rhombus, and the diagonals of a rhombus are perpendicular bisectors of each other, then the diagonals of a square must be perpendicular bisectors of each other

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A square has four-fold rotational symmetry - as you rotate it around the point where the diagonals cross, there are four positions in which it looks the same. This means that the four angles at the centre must be equal. They will each measure 360/4 = 90 degrees, so the diagonals are perpendicular. Also. the four segments joining the centre to a vertex are all equal, so the diagonals bisect each other.

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Q: Why are the diagonals of a square perpendicular bisectors of each other?
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