Giải:
Kẻ IF là tia phân giác của \(\widehat{BIC}\)
Ta có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\) ( 3 góc của \(\Delta ABC\) )
\(\Rightarrow\widehat{B}+\widehat{C}=120^o\) ( do \(\widehat{A}=60^o\) )
\(\Rightarrow\frac{1}{2}\left(\widehat{B}+\widehat{C}\right)=\frac{1}{2}.120^o\)
\(\Rightarrow\frac{1}{2}\widehat{B}+\frac{1}{2}\widehat{C}=60^o\)
\(\Rightarrow\widehat{B_1}+\widehat{C_1}=60^o\)
Trong \(\Delta BIC\) có: \(\widehat{BIC}+\widehat{B_1}+\widehat{C_1}=180^o\)
\(\Rightarrow\widehat{BIC}+60^o=180^o\)
\(\Rightarrow\widehat{BIC}=120^o\)
Vì IF là tia phân giác của \(\widehat{BIC}\) nên:
\(\widehat{I_2}=\widehat{I_3}=\frac{1}{2}\widehat{BIC}=60^o\)
Góc ngoài: \(\widehat{I_4}=\widehat{B_1}+\widehat{C_1}=60^o\)
\(\widehat{I_1}=\widehat{B_1}+\widehat{C_1}=60^o\)
Xét \(\Delta EIB,\Delta FIB\) có:
\(\widehat{I_1}=\widehat{I_2}\left(=60^o\right)\)
\(IB\): cạnh chung
\(\widehat{B_1}=\widehat{B_2}\left(=\frac{1}{2}\widehat{B}\right)\)
\(\Rightarrow\Delta EIB=\Delta FIB\left(g-c-g\right)\)
\(\Rightarrow IE=IF\) ( cạnh t/ứng ) (1)
Xét \(\Delta DIC,\Delta FIC\) có:
\(\widehat{I_3}=\widehat{I_4}\left(=60^o\right)\)
\(IC\): cạnh chung
\(\widehat{C_1}=\widehat{C_2}\left(=\frac{1}{2}\widehat{C}\right)\)
\(\Rightarrow\Delta DIC=\Delta FIC\left(g-c-g\right)\)
\(\Rightarrow ID=IF\) ( cạnh t/ứng ) (2)
Từ (1) và (2) suy ra \(ID=IE\)
\(\Rightarrowđpcm\)
