a/ Ta có:
AM = ME
BM = MC
\(\widehat{AMB}=\widehat{BME}\)
\(\Rightarrow\Delta ABM=\Delta CME\)(c.g.c)
b/
\(\Delta AMC=\Delta BME \left(c.g.c\right)\)
\(\Rightarrow\widehat{E\text{A}C}=\widehat{A\text{E}C}\)
\(\Rightarrow AC\)//\(BE\)
c/\(\Delta AMB=\Delta CME\Rightarrow\widehat{ABC}=\widehat{BCE}\)
=>\(\Delta IMB=\Delta CMK\left(c.g.c\right)\)
\(\Rightarrow\widehat{IMB}=\widehat{CMK}\)
Ta có: \(\widehat{BMI}+\widehat{IMC}=180^0\)(2 góc kề bù)
mà\(\widehat{IMB}=\widehat{CMK}\)
\(\Rightarrow\widehat{CMK}+\widehat{IMC}=180^0\)
=>\(\widehat{IMK}=180^0\)
Hay I,M,K thẳng hàng