Geometry: Prove that two angles are not equal

By contradiction: if the angles were equal, then $BM:MC=AB:AC$ (angle bisector theorem) and as a consequence $BM>MC$, which is false.

A mostly visual proof:

Construct a second square with edge $BC$ as shown. Since $AC=BD$, $CM=BM$, and $\angle ACM$ and $\angle DBM$ are right angles, $\triangle ACM \cong \triangle DBM$.

Therefore $\angle CMA \cong \angle BMD$, so $A$, $M$, and $D$ are collinear. And $m\angle BDM = m\angle CAM = \beta$.

Now if $\alpha = \beta$, then $\triangle ABD$ is isosceles, with $AB = BD$. But this is plainly false since $AB = BD \sqrt{2}$. So $\alpha \neq \beta$.

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If $\alpha=\beta$, then: $$1=\tan 45^\circ=\tan(\alpha+\beta)=\tan(2\beta)=\frac{2\tan\beta}{1-\tan^2\beta}=\frac{2\cdot \frac12}{1-\left(\frac12\right)^2}=\frac43,$$ hence a contradiction. So, $\alpha\ne \beta$.