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This question is in chapter 5 $\S$3 of Serge Lang's Basic Mathematics.

Am I wrong in thinking there is no right triangle that exists such that the length of the hypotenuse is equal to the length of a leg? It seems to me that for a right triangle, the length of the hypotenuse will always be greater than the length of a leg, but never equal.

josh h
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    Hint: Euclid's Book I Proposition 19: "in any triangle the side opposite the greater angle is greater". – dxiv Jun 12 '23 at 23:22
  • It's equal if at least one in $a$ or $b$ is $0$. – Cedric Martens Jun 12 '23 at 23:23
  • @dxiv I don't understand how your comment answers josh's question, perhaps the title of this question should be edited as josh is interested in the equality in the statement. – Cedric Martens Jun 12 '23 at 23:25
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    @CedricMartens In any right triangle, the right angle is the largest angle, so the hypotenuse is the largest side. Note that a degenerate triangle with a side equal to zero is not customarily called a right triangle. – dxiv Jun 12 '23 at 23:29
  • Is this question all about why Lang didn't say $>$ instead of $\ge$? Maybe because he doesn't need a stronger statement? Maybe because he considers degenerate triangles (those with side $0$) as well? Maybe because he has not developed the necessary tools to derive the stricter inequality by that point? I don't have the book at hand, so I really don't know - but you are right: in (what we commonly call) a right triangle, the inequality will be strict. –  Jun 13 '23 at 01:42
  • @StinkingBishop Yes, my question is indeed about why Lang wrote $\ge$ instead of $>$. – josh h Jun 13 '23 at 02:23

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An equality can happen in degenerate triangle. However, it wouldn't be a right triangle anymore.

By the pythagorean theorem,

$$c^2 = a^2 + b^2$$

If $a=0$ or $b=0$, then it is the case that the hypothenuse is the same length as a leg.

So it only is true, if we include degenerate triangles.

In the case that $a$ and $b$ must be positive, then $c^2$ is greater than $a^2$ or $b^2$.

Suppose that $c^2$ = $a^2$,

By definition, we know that $c^2 = a^2 + b^2$, and since $b^2$ is non-zero then $a^2 \lt a^2 + b^2$, meaning that $c^2$ cannot be equal to $a^2$. The same argument applies to $b^2$.

Cedric Martens
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By the Pythagorean Theorem, we know if a right triangle has legs of length $a,b$, then it's hypotenous has length $$ c:=\sqrt{a^2+b^2}. $$ Since these lengths $a,b$ are both positive, as they connect two distinct points in a plane, it follows that $$ c > \sqrt{\max\{a^2,b^2\}} = max\{a,b\}. $$

Hope this help. Thanks for the awesome question! :D

wbitting95
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