a,
Ta có:
$\widehat{AIC}=180^o-(\widehat{IAC}+\widehat{ICA})$(do $ΔIAC$)
$=180^o-(\dfrac{\widehat{ACB}}{2}+\dfrac{\widehat{CAB}}{2})$
$=180^o-(\dfrac{\widehat{ACB}+\widehat{CAB}}{2})$
$=180^o-\dfrac{180^o-\widehat{ABC}}{2}$(do $ΔABC$)
$=180^o-\dfrac{180^o-60^o}{2}$
$=120^o$
b,
Gọi $IK$ là đườn phân giác $\widehat{AIC}$
$⇒\widehat{AIK}=\widehat{CIK}=60^o=\dfrac{1}{2}\widehat{AIC}$
$⇒\widehat{AIK}=\widehat{AIN}$;$\widehat{CIK}=\widehat{CIM}$
Xét $ΔAIK$ và $ΔAIN$ có:
$\widehat{AIK}=\widehat{AIN}$
$AI$ chung
$\widehat{IAK}=\widehat{IAN}$
$⇒ΔAIK=ΔAIN(c.g.c)$
$⇒IK=IN(1)$
chứng minh tương tự ta có: $ΔCIK=ΔCIM(c.g.c)⇒IK=IM(2)$
Từ $(1)(2)⇒IM=IN$
