Giải thích các bước giải:
a.Ta có :
$\begin{cases}BI=IN\\\widehat{AIN}=\widehat{BIC}\\ IA=IC\end{cases}\rightarrow\Delta BIC=\Delta NIA(c.g.c)$
b.Từ câu a
$\rightarrow \widehat{ANI}=\widehat{IBC}\rightarrow AN//BC$
c.Chứng minh tương tự ta có :
$AM//BC\rightarrow A,M,N$ thẳng hàng
d.Vì K,I là trung điểm AB, AC
$\rightarrow\dfrac{AK}{AB}=\dfrac{AI}{AC}=\dfrac{1}{2}\rightarrow KI//BC$
Ta có :
$\begin{split}\widehat{KDI}&=180^o-\widehat{DKC}-\widehat{CKI}-\widehat{KIB}-\widehat{BID}\\&=180^o-\dfrac{1}{2}\widehat{BKC}-\widehat{KCB}-\widehat{IBC}-\dfrac{1}{2}\widehat{BIC}\\&=(90^o-\dfrac{1}{2}\widehat{BKC})-\widehat{KCI}-\widehat{IBK}-(90-\dfrac{1}{2}\widehat{BIC})\\&=\dfrac{1}{2}(180^o-\widehat{BKC})-\widehat{KCI}-\widehat{IBK}-\dfrac{1}{2}(180-\widehat{BIC})\\&=\dfrac{1}{2}(\widehat{KBC}+\widehat{KIC})-\widehat{KCI}-\widehat{IBK}-\dfrac{1}{2}(\widehat{IBC}+\widehat{ICB})\\&=\dfrac{1}{2}(2\widehat{KBI}+\widehat{KIC})-\widehat{KCI}-\widehat{IBK}-\dfrac{1}{2}(\widehat{KCI}+2\widehat{KBI})\\&=\dfrac{1}{2}(\widehat{KBI}+\widehat{KCI})\end{split}$
$\rightarrow\widehat{KBI}+\widehat{KCI}=2\widehat{KDI}$
