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So in, $\triangle ABC$ it is given that $AB^2=AC^2+BC^2$. Let us assume that $\angle C\neq{90}^{\circ}$. And let us make a perpendicular $AD$ to $BC$.
Now, by the Pythagorean Theorem, in $\triangle ADB$ & $\triangle ADC$, $AB^2=AD^2+BD^2$ and $AD^2+DC^2=AC^2$. Adding both, we get $$AB^2+AC^2=2AD^2+BD^2+DC^2$$ By, $AD^2=AC^2-DC^2$, $$AB^2+AC^2=2AC^2-2DC^2+BD^2+DC^2$$ $$AB^2-AC^2=BD^2-DC^2$$ By, $BC^2=AB^2-AC^2$ and $a^2-b^2=(a-b)(a+b)$ $$BC^2=(BD-DC)(BD+DC)$$ $$BD+DC=BD-DC$$ So, $DC=0$. So it means that $C$ and $D$ are coincident points. Which contradicts the fact that $\angle C\neq90^{\circ}$. Thus, proved by contradiction.
I know this is so easy proof, but what happened is that my teacher said that this proof is not in the book, so it is wrong. He didn't give me any marks in this question. Also, I didn't remember my book's proof that day, so I made my own and wrote. But it seems alright to me. So, is there really any mistake?

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    $\begingroup$ Your proof looks correct to me. @SilviaGhinassi: I don't think that he ever claims that it does. $\endgroup$
    – Dylan
    Sep 1, 2015 at 15:02
  • $\begingroup$ @Dylan he edited the post adding passages. I guess I am wrong, thanks $\endgroup$ Sep 1, 2015 at 15:06
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    $\begingroup$ So.....your teacher believes that this book contains every correct argument? That must be an amazing book --- or a very deluded teacher. $\endgroup$
    – WillO
    Sep 1, 2015 at 15:11
  • $\begingroup$ I think the second part is valid. What can I do? $\endgroup$ Sep 1, 2015 at 15:13
  • $\begingroup$ Speak to your teacher? Speak to another teacher or the head of department? How important is it? How far is it worth fighting? $\endgroup$
    – Dylan
    Sep 1, 2015 at 17:40

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Your proof is nice but incomplete, since you are assuming $\angle C<90^\circ$ (that is point $D$ falling inside $BC$). To complete it, you should also examine the $\angle C>90^\circ$ case. However, if I were your teacher I would reward your effort with a fair mark instead of dismissing it.

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  • $\begingroup$ I think that his proof, as written, works in the $\angle C > 90^\circ$ case as well, except that the justification that allows you to go from the second last line of algebra to the last line is now $BC=BD-DC$ instead of $BC=BD+DC$. It should potentially however, as you say, be stated that both cases arise. $\endgroup$
    – Dylan
    Sep 1, 2015 at 17:57
  • $\begingroup$ You should also point out that $AC$ is chosen to be shorter than $BC$. $\endgroup$ Sep 1, 2015 at 19:25
  • $\begingroup$ Sorry, I forgot to wrote that case. I had written it in my test paper. I hadn't ignored it. $\endgroup$ Sep 2, 2015 at 9:59

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