Validity of my proof by contradiction of converse of Pythagorean Theorem. 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 $\triangle ADB$ &amp; $\triangle ADC$, $AB^2=AD^2+BD^2$ 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?
Your proof is nice but incomplete, since you are assuming $\angle C&lt;90^\circ$ (that is point $D$ falling inside $BC$). To complete it, you should also examine the $\angle C&gt;90^\circ$ case. However, if I were your teacher I would reward your effort with a fair mark instead of dismissing it.
Your proof is wrong as you said angle $C$ is not equal to $90^{\circ}$. But
when you derived $AB^2-AC^2$, you took it as $BC^2$, and so took $C$ as $90^{\circ}$, thereby contradicting your own assumption, hence you got $DC=0$.