Larger side / larger angle theorem:
Prop. I.18: In any triangle, the greater side is opposite the greater angle.
Proof: Assume not. Let angle $\angle ACB$ be greater than $\angle BAC$ and assume side AB is less than base BC. Extend side BA past A to X so that BX equals BC. Then in triangle BXC, sides BX and BC are equal so angles $\angle BCX = \angle BXC$. Now $\angle BAC$ is exterior to $\angle BXC$ hence $\angle BAC$ is greater. But $\angle BAC$ was assumed smaller than $\angle ACB = \angle BCX=\angle BXC$, a contradiction.
I just can't see how angle $\angle ACB = \angle BCX$. After I draw the diagram, it looks $\angle ACB$ is inside $\angle BCX$, therefore smaller than $\angle BCX$.
Your picture is not good because the angles all look the same. At least start by drawing a triangle where one side is visually much larger than the other sides. Also I think you have a typo: ACB should be less than BCX since it's an interior angle. This will give you a contradiction that BAC> BXC> ACB whereas you assumed BAC < ACB.