Ta có
\(\widehat{ABC}+\widehat{ACB}=180^o-\widehat{A}=180^o-60^o=120^0\)
\(\widehat{EBI}=\widehat{KBI}=\frac{\widehat{ABC}}{2};\widehat{DCI}=\widehat{KCI}=\frac{\widehat{ACB}}{2}\)
\(\Rightarrow\widehat{KBI}+\widehat{KCI}=\frac{\widehat{ABC}}{2}+\frac{\widehat{ACB}}{2}=\frac{\widehat{ABC}+\widehat{ACB}}{2}=\frac{120^o}{2}=60^o\)
Xét tg BIC có
\(\widehat{BIC}=180^o-\left(\widehat{KBI}+\widehat{KCI}\right)=180^o-60^o=120^o\)
\(\Rightarrow\widehat{BIE}=\widehat{CID}=60^o\) (Cùng bù với góc \(\widehat{BIC}\) )
Xét tg BIE và tg BIK có
\(\widehat{EBI}=\widehat{KBI}\)
BE=BK; BI chung
\(\Rightarrow\Delta BIE=\Delta BIK\left(c.g.c\right)\Rightarrow\widehat{BIE}=\widehat{BIK}=60^o\)
\(\Rightarrow\widehat{CIK}=\widehat{BIC}-\widehat{BIK}=120^o-60^o=60^o\)
Xét tg CIK và tg CID có
\(\widehat{DCI}=\widehat{KCI};\widehat{CID}=\widehat{CIK}=60^o\)
CI chung
\(\Rightarrow\Delta CIK=\Delta CID\left(g.c.g\right)\Rightarrow CD=CK\)
Vậy BE=BK và CD=CK nên BE+CD=BK+CK=BC (dpcm)
