Giải thích các bước giải:
a.Xét $\Delta ACE,\Delta BCD$ có:
$CA=CD$ vì $\Delta ACD$ đều
$\widehat{ACE}=180^o-\widehat{ECB}=120^o=180^o-\widehat{DCA}=\widehat{BCD}$
$CE=CB$ vì $\Delta CBE$ đều
$\to\Delta ACE=\Delta DCB(c.g.c)$
$\to\widehat{BDC}=\widehat{CAE},\widehat{AEC}=\widehat{DBC}$
$\to\widehat{IDC}=\widehat{IAC},\widehat{IEC}=\widehat{IBC}$
$\to ADIC, BCIE$ nội tiếp
b.Ta có:
$\widehat{AIB}=180^o-\widehat{IAB}-\widehat{IBA}$
$\to\widehat{AIB}=180^o-\widehat{BDC}-\widehat{DBC}$
$\to\widehat{AIB}=\widehat{DCB}$
$\to\widehat{AIB}=180^o-\widehat{DCA}=180^o-60^o=120^o$
Lại có $\Delta ABF$ đều
$\to\widehat{AIB}+\widehat{AFB}=120^o+60^o=180^o$
$\to AIBF$ nội tiếp
$\to\widehat{AIF}=\widehat{ABF}=60^o=\widehat{ADC}=\widehat{AIC}$
$\to I, C, F$ thẳng hàng
