$\Delta$ ABC đều có $\widehat{B}= \widehat{A}= 60^o$
M nằm trên AB nên CM nằm giữa CB, CA
=> $\widehat{BCM} < \widehat{BCA}= 60^o$
$\Delta$ BCM có $\widehat{B}= 60^o$, $\widehat{BCM}<60^o$
=> $\widehat{BMC}>60^o$
=> $\widehat{BCM} < \widehat{B} < \widehat{BMC}$
=> BM < MC < BC