Giải thích các bước giải:

a.Ta có:

$\widehat{BIC}=180^o-(\widehat{IBC}+\widehat{ICB})$ 

$\to\widehat{BIC}=180^o-(\dfrac12\hat B+\dfrac12\hat C)$

$\to\widehat{BIC}=180^o-\dfrac12(\hat B+\hat C)$

$\to\widehat{BIC}=180^o-\dfrac12(180^o-\hat A)$

$\to\widehat{BIC}=115^o$

b.Kẻ $KD\perp AB, KE\perp AC, D,E\in$ tia đối của tia $BA, CA$

   Kẻ $KF\perp BC, F\in CB$

Vì $BK$ là phân giác góc ngoài tại đỉnh $B$

$\to\widehat{IBK}=\widehat{IBC}+\widehat{CBK}=\dfrac12\widehat{ABC}+\dfrac12\widehat{DBC}=\dfrac12(\widehat{ABC}+\widehat{DBC})=\dfrac12\widehat{ABD}=\dfrac12\cdot 180^o=90^o$

Tương tự $\widehat{ICK}=90^o$

$\to \widehat{BKC}=180^o-\widehat{KBC}-\widehat{KCB}$

$\to \widehat{BKC}=180^o-(\widehat{IBK}-\widehat{IBC})-(\widehat{ICK}-\widehat{ICB})$

$\to \widehat{BKC}=180^o-\widehat{IBK}+\widehat{IBC}-\widehat{ICK}+\widehat{ICB}$

$\to\widehat{BKC}=180^o+(\widehat{IBC}+\widehat{IBC})-(\widehat{IBK}+\widehat{ICK})$

$\to \widehat{BKC}=180^o+(180^o-\widehat{BIC})-180^o$

$\to \widehat{BKC}=180^o+180^o-\widehat{BIC}-180^o$

$\to \widehat{BKC}=180^o-\widehat{BIC}$

$\to \widehat{BKC}=65^o$

c.Vì $BI, CI$ là phân giác $\Delta ABC\to AI$ là phân giác $\hat A$

Ta có: $KB$ là phân giác $\widehat{CBD}, KF\perp BC, KD\perp BD$

$\to KD=KF$

Tương tự $KF=KE$

$\to KD=KE$

Ta có: $KD\perp AB, KE\perp AC, KD=KE$

$\to K\in$ phân giác $\widehat{BAC}$

$\to K\in AI$

$\to A, I, K$ thẳng hàng