<s>[INST] Geometry: Prove that two angles are not equal (This is just a question for fun. I saw a commercial logo today and I was inspired.
I have posted answers for this question and you may post alternative answers!)

Question
In the figure, $\triangle ABC$ is half of a square and $M$ is the midpoint of $BC$. Prove that $\alpha\neq\beta$.


Solution

 $\triangle ABM$ and $\triangle AMC$ have the same area. They have a
 common side $AM$.

 Note that the area of either triangle is given by
 $S=\frac12(AB)(AM)\sin\alpha=\frac{1}{2}(AC)(AM)\sin\beta$.

 But $AB\neq AC$. So the equality holds only if $\alpha\neq\beta$.

 [/INST] First assume that $\alpha$ = $\beta$ and that both would be equal to $\frac{\pi}{8}^\circ$ (since  $\alpha = \beta$ and $\alpha + \beta = \frac{\pi}{4} \to 2\beta = \frac{\pi}{4} \to \beta = \frac{\pi}{8}$).
If we call side $\overline{AC}$, n,  then side $\overline{MC}$ is $\frac{n}{2}$. Next, we solve for $\overline{AM}$ in terms of n to show that $\alpha = \beta$ for all values n. 
$\overline{AM}^2 = n^2 + (\frac{n}{2})^2 = n^2 + \frac{n^2}{4} = \frac{4n^2 + n^2}{4} = \frac{5n^2}{4}$
$\overline{AM} = \sqrt{\frac{5n^2}{4}} = \frac{\sqrt{5n^2}}{2} = \frac{n\sqrt{5}}{2}$ by the pythagorean theorem
Now that we have $\overline{AM}$, we can find $cos{\beta}$.
$\cos{\beta}=\frac{n}{\overline{AM}} = \frac{n}{\frac{n\sqrt{5}}{2}} = \frac{2n}{n\sqrt{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$
$\cos{\beta} = \frac{2\sqrt{5}}{5}$
We have that $\cos{\beta} = \frac{2\sqrt{5}}{5}$ and we already established that $\beta = \frac{\pi}{8}$.
This would mean that  $\cos{\frac{\pi}{8}} = \frac{2\sqrt{5}}{5}$. We can find $\cos{\frac{\pi}{8}} = \frac{\sqrt{2 + \sqrt{2}}}{2}$, which implies that $\frac{\sqrt{2 + \sqrt{2}}}{2} = \frac{2\sqrt{5}}{5}$.
If we square both sides, we get that $\frac{2+\sqrt{2}}{4} = \frac{20}{25} = \frac{4}{5}$. Then we can split the fraction, simplify, then subtract and simplify.
$\frac{2}{4} + \frac{\sqrt{2}}{4} = \frac{4}{5}$
$\frac{1}{2} + \frac{\sqrt{2}}{4} = \frac{4}{5}$
$\frac{\sqrt{2}}{4} = \frac{4}{5} - \frac{1}{2} $
$\frac{\sqrt{2}}{4} = \frac{3}{10}$
Next we can multiply both sides by 4 and conclude that...
$\sqrt{2} = \frac{12}{10} = \frac{6}{5}$
This is impossible since it has been proven that $\sqrt{2}$ is irrational and can't be written as a ratio of 2 integers. This would mean that $\sqrt{2} \neq \frac{6}{5}$ or $\frac{\sqrt{2 + \sqrt{2}}}{2} \neq \frac{2\sqrt{5}}{5}$ or $\cos{\frac{\pi}{8}}\neq \frac{2\sqrt{5}}{5}$
This means that $\beta \neq \frac{\pi}{8}$. Since $\alpha + \beta = \frac{\pi}{4}$, $\beta$ would have to be equal to $\frac{\pi}{8}$, but we proved that it cant be. This means that $\alpha \neq \beta$ since $\beta$ can no longer be multiplied by 2 to be $\frac{\pi}{4}$
This method works with tangent and sine as well (I believe it's a lot less work if I had used tangent instead of cosine, but this was how I first solved it)
I'm a high-school student and this is my first response, feedback is always appreciated!
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