Sửa đề: Trên tia đối của tia $IH$ lấy điểm $K$ sao cho $HI = IK$
Lời giải:
Xét $\triangle HBI$ và $\triangle KCI$ có:
$\begin{cases}BI = IC = \dfrac12BC\quad (gt)\\\widehat{HIB} = \widehat{KIC}\quad \text{(đối đỉnh)}\\HI = IK =\dfrac12HK\quad (gt)\end{cases}$
Do đó: $\triangle HBI = \triangle KCI\ (c.g.c)$
$\Rightarrow \begin{cases}\widehat{HBI} = \widehat{KCI}\quad \text{(hai góc tương ứng)}\\HB = KC\quad \text{(hai cạnh tương ứng)}\end{cases}$
$\Leftrightarrow \widehat{HBA} + \widehat{ABC} = \widehat{KCI}$
$\Leftrightarrow \widehat{HAB} + \widehat{ABC} = \widehat{KCI}$
Ta có:
$\quad \widehat{KCF} = 360^\circ - (\widehat{ACF} + \widehat{ACB} + \widehat{KCI}$
$\Leftrightarrow \widehat{KCF} = 360^\circ - (60^\circ - 180^\circ - \widehat{ABC} - \widehat{BAC} + \widehat{HAB} + \widehat{ABC})$
$\Leftrightarrow \widehat{KCF} = 120^\circ + \widehat{BAC} - \widehat{HAB}$
$\Leftrightarrow \widehat{KCF} = 60^\circ + 60^\circ + \widehat{BAC} - \widehat{HAB}$
$\Leftrightarrow \widehat{KCF} = \widehat{CAF} + 2\widehat{HAB} + \widehat{BAC} - \widehat{HAB}$
$\Leftrightarrow \widehat{KCF} = \widehat{CAF} + \widehat{BAC} + \widehat{HAB}$
$\Leftrightarrow \widehat{KCF} = \widehat{HAF}$
Xét $\triangle KCF$ và $\triangle HAF$ có:
$\begin{cases}CF = AF\quad \text{($\triangle ACF$ đều)}\\ \widehat{KCF} = \widehat{HAF}\quad (cmt)\\KC = HA\quad (=HB)\end{cases}$
Do đó: $\triangle KCF = \triangle HAF\ (c.g.c)$
b) Ta có:
$\triangle KCF = \triangle HAF$ (câu a)
$\Rightarrow \widehat{KFC} = \widehat{AFH}$ (hai góc tương ứng)
mà $\widehat{AFH} + \widehat{CFH} = \widehat{AFC} = 60^\circ$
nên $\widehat{KFC} + \widehat{CFH} = 60^\circ$
hay $\widehat{KFH} = 60^\circ$
Xét $\triangle KHF$ có:
$KF = HF\quad (\triangle KCF = \triangle HAF)$
$\widehat{KFH} = 60^\circ\quad (gt)$
Do đó: $\triangle KHF$ đều
