Assume that $ \widehat{ACB}>90^0$ and $ \widehat{CAB}\leq 45^0$. Contruct an $ E$-isosceles right-angled triangle $ AED$ with $ AD=\frac{\sqrt{2}+1}{2}AB$ and $ C,E$ are same side with $ AB$. Then $ AD\leq \sqrt{2}+1$ because $ AB\leq 2$ We prove that: $ C$ is inside triangle $ AED$. In fact: Call $ M$ is midpoint of $ AB$ and $ F$ is foot of peperdicular line from $ M$ to $ DE$. Then $ MF=\frac{MD}{\sqrt{2}}=\frac{AD-AM}{\sqrt{2}}=\frac{AB}{2}=AM=MB$ It mean $ DE$ is tangent with $ (AFB)$. But $ C$ is inside $ (AFB)$, it mean $ C$ is inside triangle $ AED$.