a) $\widehat{A}+\widehat{B}+\widehat{C}=\frac{180^{\circ}}{2}=90^{\circ}$
b) Trong tam giác BOC ta có:
$\widehat{BOC}=180^{\circ}-\frac{\widehat{B}+\widehat{C}}{2}=180^{\circ}-45^{\circ}=135^{\circ}$
a) $\widehat{A}+\widehat{B}+\widehat{C}=\frac{180^{\circ}}{2}=90^{\circ}$
b) Trong tam giác BOC ta có:
$\widehat{BOC}=180^{\circ}-\frac{\widehat{B}+\widehat{C}}{2}=180^{\circ}-45^{\circ}=135^{\circ}$