Prove that the diagonals are the longest lines in a square. I just pondered this question and have tried out several methods to solve it(mainly using trigonometry). However, I am not satisfied with my trigonometrical proof and is looking for better proofs. Give an _elegant proof_ that the diagonals are the longest lines in a square. It would rather be nice if the proof is by **"reductio ad absurdum"** method. Thank you :). Bonus : Also prove that the diagonals of a rectangle are the longest lines.

It is sufficient to consider the two cases (since others are repetetive): without loss of generality, take two arbitrary points on two sides such that $0\le x\le y\le a$:

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Case 1: From the Pythagoras: $$\sqrt{x^2+y^2}\le \sqrt{a^2+a^2}=d.$$ Case 2: From the Pythagoras: $$\sqrt{a^2+(y-x)^2}\le \sqrt{a^2+a^2}=d,$$because $y-x$ is maximum, when $y=a$ and $x=0$.