Giải thích các bước giải:

a.Xét $\Delta CNB,\Delta CIA$ có:

$\widehat{CBN}=90^o-\widehat{CBA}=\widehat{CAB}=\widehat{CAI}$

$\widehat{BCN}=90^o-\widehat{ICB}=\widehat{ICA}$

$\to \Delta CBN\sim\Delta CAI(g.g)$

b.Từ a

$\to \dfrac{CN}{CI}=\dfrac{CB}{CA}$

Mà $\widehat{ICN}=\widehat{ACB}(=90^o)$

$\to \Delta CNI\sim\Delta CBA(c.g.c)$

c.Từ b $\to \widehat{CNI}=\widehat{CBI}$

$\to \widehat{INM}=\widehat{ACB}$

Tương tự $\widehat{IMN}=\widehat{BAC}$

$\to \Delta IMN\sim\Delta CBA(g.g)$

$\to \widehat{MIN}=\widehat{BCA}=90^o$

d.Kẻ $CD\perp AB$

Ta có:

$\dfrac{S_{MIN}}{S_{ABC}}=2$

$\to \dfrac{\dfrac12IC.MN}{\dfrac12CD.AB}=2$

$\to \dfrac{IC.MN}{CD.AB}=2$

Từ c $\to \dfrac{S_{MIN}}{S_{ABC}}=\dfrac{MN^2}{AB^2}$

$\to \dfrac{MN^2}{AB^2}=2$

$\to \dfrac{MN}{AB}=\sqrt2$

$\to \dfrac{IC}{CD}.\sqrt2=2$

$\to \dfrac{IC}{CD}=\sqrt2$

$\to \dfrac{CD}{IC}=\dfrac1{\sqrt2}$

$\to \sin\widehat{DIC}=\dfrac1{\sqrt2}$

$\to \widehat{CID}=45^o$

$\to \widehat{BIC}=45^o$