\(BC=\sqrt{AB^2+AC^2}=\sqrt{2AB^2}=AB\sqrt{2}\\ \Rightarrow\dfrac{BC}{AB}=\sqrt{2}\\ \Rightarrow\dfrac{BC}{AB}=\dfrac{PB}{PA}=\dfrac{PC}{PB}=\sqrt{2}\)
Lại có \(\dfrac{PA}{AB}=\dfrac{1}{\dfrac{BC}{\sqrt{2}}}=\dfrac{\sqrt{2}}{BC}=\dfrac{PB}{BC}\left(PB=\sqrt{2}\right)\)
Do đó \(\Delta APB\sim\Delta BPC\left(c.g.c\right)\)
\(\Rightarrow\widehat{PBC}=\widehat{PAB}\)
\(\Rightarrow\widehat{APB}=180^0-\left(\widehat{PAB}+\widehat{PBA}\right)=180^0-\left(\widehat{PBC}+\widehat{PBA}\right)=180^0-\widehat{ABC}=180^0-45^0=135^0\)
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