Giải thích các bước giải:
a) Xét tam giác EBC có: \(\widehat{BEC}\) là góc ngoài tam giác EAC
Khi đó: $\widehat{BEC}=\widehat{A}+\widehat{ECA}=90^o+\widehat{ECA}>90^o$
Suy ra đpcm.
b) Đặt \(\widehat{B}=x\), ta có:
\(\widehat{B}+\widehat{BCA}=90^{\circ}\) (do \(\triangle{ABC}\) vuông tại \(A\))
\(\Rightarrow \widehat{BCA}=90^{\circ}- x\Rightarrow \widehat{C_1}=\widehat{C_2}=\frac{\widehat{BCE}}{2}=\frac{90^{\circ}- x}{2}\)
Xét \(\triangle{BEC}\), có: \(\widehat{B}+\widehat{BEC}+\widehat{C_1}=180^{\circ}\)
\(\Rightarrow x+110^{\circ}+\frac{90^{\circ}- x}{2}=180^{\circ}\)
\(\Rightarrow \frac{90^{\circ}- x+2x}{2}=70^{\circ}\)
\(\Rightarrow 90^{\circ}+x=140^{\circ} \Rightarrow x=50^{\circ}\)
Vậy \(\widehat{B}=50^{\circ}\)
