Giải thích các bước giải:
a.Ta có:
$\begin{split}\widehat{BID}&=\dfrac{1}{2}\widehat{BIC}\\&=\dfrac{1}{2}(180^o-\widehat{BCI}-\widehat{IBC})\\&=\dfrac{1}{2}(180^o-\dfrac{1}{2}\widehat{BCA}-\dfrac{1}{2}\widehat{ABC})\\&=\dfrac{1}{2}(180^o-\dfrac{1}{2}(\widehat{BCA}+\widehat{ABC})\\&=\dfrac{1}{2}(180^o-\dfrac{1}{2}(180^o-\widehat{BAC})\\&=60^o\end{split}$
Lại có :
$\begin{split}\widehat{NIB}&=\widehat{IBC}+\widehat{ICB}\\&=\dfrac{1}{2}\widehat{ABC}+\dfrac{1}{2}\widehat{ACB}\\&=\dfrac{1}{2}(\widehat{ABC}+\widehat{ACB}&=\dfrac{1}{2}(180^o-\widehat{BAC})\\&=60^o\end{split}$
$\rightarrow\widehat{NIB}=\widehat{BID}$
$\rightarrow\Delta NIB=\Delta DIB(g.c.g)$
$\rightarrow BN=BD$
b.Chứng minh tương tự câu a
$\rightarrow CD=CM$
$\rightarrow BN+CM=BD+CD=BC\rightarrow đpcm$
