Prove that in $\triangle ABC$ if $\gamma > \alpha$ then $|AB| > |BC|$

Question

For $\alpha$ being an angle with vertex $A$, $\beta$ with vertex $B$ and $\gamma$ with vertex $C$.

Yeah, I know, it's a pretty basic theorem and I could look it up on Wikipedia, but currently I'm in the process of learning how to write and comprehend mathematical proofs, so this will be a good example for me, regardless of the difficulty (actually, I think it's even better to practice on basic proofs). So yeah, don't shy away from lengthy explanation, it will surely be of great help to me. Thank you in advance!

Answer 1

Note that $\displaystyle \frac{|AB|}{|BC|}=\frac{\sin \gamma}{\sin\alpha}$.

If $\displaystyle \frac{\pi}{2}>\gamma>\alpha$, then $\sin\gamma>\sin\alpha$ and hence $|AB|>|BC|$.

If $\displaystyle \gamma>\frac{\pi}{2}$, then $\displaystyle \frac{\pi}{2}>\alpha+\beta>\alpha$ and $\sin\gamma=\sin(\pi-\alpha-\beta)=\sin(\alpha+\beta)>\sin\alpha$ and hence $|AB|>|BC|$.

Answer 2

Inside the angle $\gamma$ at $C$ copy the angle $\alpha$. Let this line intersect $AB$ at $D$. Then you have $$\alpha=\angle ACD$$ and by isosceles triangles you have $$|AD|=|CD|$$ this implies that $$|AB|=|CD|+|DB|$$ so your statement becomes $$|CB|< |CD|+|DB|$$ and this is just the assertion that one side of a triangle is less that the sum of the other two sides.