Ta có:
$\widehat{ABC}$ + $\widehat{ACB}$ + $\widehat{BAC}$ = $180^o$ (đ/lí)
⇒ $\widehat{ABC}$ + $\widehat{ACB}$ = $180^o$ - $\widehat{BAC}$ = $180^o$ - $70^o$ = $110^o$
Vì BD và CE là 2 đường phân giác của $\widehat{ABC}$ và $\widehat{ACB}$ nên ta có:
$\widehat{CBI}$ + $\widehat{BCI}$ = $\frac{1}{2}$ $\widehat{ABC}$ + $\frac{1}{2}$ $\widehat{ACB}$
= $\frac{1}{2}$ ( $\widehat{ABC}$ + $\widehat{ACB}$ )
= $\frac{1}{2}$ . $110^o$ = $55^o$
Ta có:
$\widehat{CBI}$ + $\widehat{BCI}$ + $\widehat{BIC}$ = $180^o$ (đ/lí)
⇒ $\widehat{BIC}$ = $180^o$ - ($\widehat{CBI}$ + $\widehat{BCI}$) = $180a^o$ - $55^o$ = $125^o$
Vậy $\widehat{BIC}$ = $125^o$
