Question: User: Given: $\triangle ABC$ is not a right $\triangle$\n\nProve: $a^2 + b^2 \neq c^2$\n\nNOTE: $AB = c$, $AC = b$, and $CB = a$\n\nGraph: A triangle labeled $A$, $B$, and $C$ is shown in the bottom-left area; $A$ is at the upper-left, $C$ is at the lower-left, and $B$ is at the lower-right, with side $CB$ drawn horizontally, side $AC$ slanting down from $A$ to $C$, and side $AB$ slanting down from $A$ to $B$. position_hint: bottom-left\n\nWhat mathematical theorem does this image come from? How can I prove this using a statements and results table.
1. **Problem Statement:**\nGiven a triangle $\triangle ABC$ where $AB = c$, $AC = b$, and $CB = a$, and the triangle is not a right triangle. We need to prove that $a^2 + b^2 \neq c^2$.\n\n2. **Relevant Theorem:**\nThe Pythagorean theorem states that in a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides: $$c^2 = a^2 + b^2.$$\nSince $\triangle ABC$ is not right-angled, this equality does not hold.\n\n3. **Important Rule:**\nFor any triangle, the Law of Cosines applies: $$c^2 = a^2 + b^2 - 2ab \cos(\gamma),$$ where $\gamma$ is the angle opposite side $c$.\n\n4. **Proof Using Law of Cosines:**\nSince $\triangle ABC$ is not right-angled, $\gamma \neq 90^\circ$. Therefore, $\cos(\gamma) \neq 0$.\n\n5. Substitute into the Law of Cosines formula:\n$$c^2 = a^2 + b^2 - 2ab \cos(\gamma).$$\n\n6. Rearranging to compare $a^2 + b^2$ and $c^2$:\n$$a^2 + b^2 - c^2 = 2ab \cos(\gamma).$$\n\n7. Since $\cos(\gamma) \neq 0$, the right side is not zero, so:\n$$a^2 + b^2 - c^2 \neq 0,$$\nwhich implies\n$$a^2 + b^2 \neq c^2.$$\n\n8. **Statements and Reasons Table:**\n| Statement | Reason |\n|---|---|\n| $\triangle ABC$ is not right-angled | Given |\n| $c^2 = a^2 + b^2 - 2ab \cos(\gamma)$ | Law of Cosines |\n| $\gamma \neq 90^\circ$ so $\cos(\gamma) \neq 0$ | Definition of right angle |\n| $a^2 + b^2 - c^2 = 2ab \cos(\gamma)$ | Algebraic rearrangement |\n| $a^2 + b^2 \neq c^2$ | Since $2ab \cos(\gamma) \neq 0$ |\n\n**Final conclusion:** For a non-right triangle, $a^2 + b^2 \neq c^2$.