Giải:
a) Xét \(\Delta ADB,\Delta ADC\) có:
\(AB=AC\left(gt\right)\)
\(IB=IC\left(=\frac{1}{2}BC\right)\)
\(AI\): cạnh chung
\(\Rightarrow\Delta ADB=\Delta ADC\left(c-c-c\right)\)
\(\Rightarrow\widehat{ADB}=\widehat{ADC}\) ( cạnh tương ứng )
b) Ta có: \(\widehat{ADB}+\widehat{ADC}=180^o\) ( kề bù )
Mà \(\widehat{ADB}=\widehat{ADC}\)
\(\Rightarrow\widehat{ADB}=\widehat{ADC}=90^o\)
hay \(AD\perp BC\)
c) Vì \(\Delta ADB=\Delta ADC\)
\(\Rightarrow\widehat{A_1}=\widehat{A_2}\) ( 2 góc tương ứng )
\(\Rightarrow\) AD là tia phân giác của \(\widehat{BAC}\)
\(\Rightarrow\widehat{BAD}=\widehat{DAC}=\frac{1}{2}\widehat{BAC}=40^o\)
Vì \(\Delta ADB=\Delta ADC\)
\(\Rightarrow\widehat{B}=\widehat{C}\)
Ta có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\) ( các góc trong \(\Delta ABC\) )
\(\Rightarrow80^o+\widehat{B}+\widehat{C}=180^o\)
\(\Rightarrow\widehat{B}+\widehat{C}=100^o\)
Mà \(\widehat{B}=\widehat{C}\)
\(\Rightarrow\widehat{B}=\widehat{C}=50^o\)
Vậy...
