Let $A B C D$ be a square, $P$ inside $A B C D$ such that $P A=1, P B=2, P C=3$. Calculate $\widehat{A P B}$.
Let $\rho$ be the rotation centered at $B$ with an angle of $90^{\circ}$. It maps $C$ to $A$, fixes $B$, and maps $P$ to a point $P^{\prime}$ such that $\widehat{P B P^{\prime}}=90^{\circ}$ and $P^{\prime} B=P B=2$. Additionally, $P^{\prime} A=P C=3$ (isometry). According to the Pythagorean theorem in $P B P^{\prime}$, $P P^{\prime}=2 \sqrt{2}$, so by the converse of the Pythagorean theorem, the triangle $P A P^{\prime}$ is right-angled at $P$. Since $P B P^{\prime}$ is an isosceles right triangle at $B$, we finally have $\widehat{A P B}=\widehat{A P P^{\prime}}+\widehat{P P^{\prime} B}=90^{\circ}+45^{\circ}=135^{\circ}$.