Cho

- Hướng dẫn giải

Kẻ IH là tia phân giác $\widehat{BIC}$

Ta có: $\widehat{CBD}=\widehat{ABD}=\frac{1}{2}\widehat{ABC}$ (BD là tia phân giác )

 $\widehat{BCE}=\widehat{ACE}=\frac{1}{2}\widehat{ACB}$ (CE là tia phân giác )

Mà $\widehat{BAC}+\widehat{ABC}+\widehat{ACB}=180°$(định lí tổng 3 góc trong tam giác)

$\begin{array}{l}\Rightarrow \widehat{ABC}+\widehat{ACB}=180°-\widehat{BAC}=180°-60°=120°\\ \Rightarrow \widehat{CBD}+\widehat{BCE}=\frac{1}{2}\left(\widehat{ABC}+\widehat{ACB}\right)=\frac{1}{2}.120°=60°\end{array}$

$\Delta BIC$ có: $\widehat{BIC}=180°-\left(\widehat{CBD}+\widehat{BCE}\right)=180°-60°=120°$

$\Rightarrow \widehat{BIH}=\widehat{CIH}=\frac{1}{2}\widehat{BIC}=60°$(IH là tia phân giác $\widehat{BIC}$)

$\widehat{BIE}=180°-\widehat{BIC}=180°-120°=60°$

Có: $\widehat{BIE}=\widehat{CID}=60°$(2 góc đối đỉnh)

Xét $\Delta BIE$ và $\Delta BIH$ có:

$\left.\begin{array}{l}\widehat{BIE}=\widehat{BIH}=60°\\ BIchung\\ \widehat{EBI}=\widehat{HBI}\left(\widehat{ABD}=\widehat{CBD}\right)\end{array}\right\}\Rightarrow \Delta BIE=\Delta BIH\left(g.c.g\right)$

 IE = IH (2 cạnh tương ứng)

Xét $\Delta DIC$ và $\Delta HIC$ có:

$\left.\begin{array}{l}\widehat{DIC}=\widehat{HIC}=60°\\ ICchung\\ \widehat{ICH}=\widehat{ICD}\left(\widehat{BCE}=\widehat{ACE}\right)\end{array}\right\}\Rightarrow \Delta DIC=\Delta HIC\left(g.c.g\right)$

$\Rightarrow \left.\begin{array}{r}ID = IH\\ Mà IE = IH\end{array}\right\}$ => ID = IE (đpcm)