Longest line in a rectangle

Question

Given a rectangle, how can one show that the euclidian distance between any two points inside the rectangle (or on its borderies) is smaller or equal than the distance of the diagonal of that rectangle? I tried to think of a geometric proof, but failed... any help is appreciated. Thanks in advance.

Answer 1

You need an easy lemma: if $P$ is a point inside an isosceles triangle $AOB$ of vertex $O$, then $OP\le OA$.

PROOF. Produce $OP$ to meet base $AB$ at $Q$. One of $\angle OQA$ or $\angle OQB$ is not acute, for instance $\angle OQA$. It follows that $\angle OQA>\angle QAO$, whence $OA\ge OQ\ge OP$, QED.

If $ABCD$ is a rectangle with center $O$ and $PQ$ a segment in it, by the above lemma we have $PQ\le OP+OQ\le 2OA$, as it was to be proved.

Answer 2

This a sort of geometrical proof, maybe. Draw a point (c1) in the centre of the rectangle. Rotate the rectangle 360 degrees. The 4 vertices which are the limits of the main diagonals (of length L) trace a circle of radius R1 = L/2. Any other points on the perimeter of the rectangle or within the rectangle trace a smaller circle. Draw a new line with endpoints thar either inside the rectangle or on the perimeter. Draw another point (c2) at the centre of the new line. Now translate the circle so that c1 lies on c2. A translation by definition does not alter the shape of the object or the spatial separation of any two points within the shape. Now rotate the circle again so that the diagonal is parallel to the new line and now lies on the new line. A rotation does not alter the shape or spatial separation of any two points within the shape. Since the endpoints of the new line were inside or on the rectangle and all points of the rectangle are either inside or on circle C1, the endpoints of the new line must also lie on the diameter of a circle that has a radius smaller than or equal to R1, so the new line must be smaller than or equal to the main rectangle diagonal.