The proof follows from the picture below. |IH|+|IF| <= |HF| and |IE|+|IG| <= |EG| and the equality holds exactly when I and K coincide.

Projecting I to the sides of the square, we get the [orthodiagonal] quadrangle ABCD, which has equal and orthogonal diagonals. The previous results translates: From all such orthodiagonals the one, having minimal perimeter has its vertices at the middles of the sides of the square (and is a square). Notice that the orthodiagonal quadrilaterals ABCD have all the same area (half the square area).

Notice the analogous property holding for even-sided convex polygons ABCD..., whose diagonals intersect at a point K. s(I) = |IA|+|IB|+... etc. takes its minimum value at K.

What happens with even sided polygons with non concurrent diagonals? What happens with odd-sided polygons and more general with arbitrary polygons? Which is the place of I, where s(I) = |IA|+|IB|+ ... takes its minimum value, generally?

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