Question

Answer 1

ΔABC vuông cân tại A
\(=>\dfrac{BC}{AB}=\dfrac{PB}{PA}=\dfrac{PC}{PB}=\sqrt{2}\)
=> ΔPBC ∼ ΔPAB
\(=>\widehat{PBC}=\widehat{PAB}\)
\(=>\widehat{APB}=180^o-\left(\widehat{PAB}+\widehat{PBA}\right)\)
\(=180^o-\left(\widehat{PBC}+\widehat{PBA}\right)\)
\(=180^o-\widehat{ABC}=180^o-45^o=135^o\)