Giải thích các bước giải:
a.Xét $\Delta CAI,\Delta CBN$ có:
$\widehat{ACI}=90^o-\widehat{ICB}=\widehat{BCN}$
$\widehat{AIC}=180^o-\widehat{BIC}=360^o-90^o-90^o-\widehat{BIC}=360^o-\widehat{IBN}-\widehat{ICN}=\widehat{BNC}$
$\to \Delta CAI\sim\Delta CBN(g.g)$
b.Gọ $AB\cap MN=D$
Xét $\Delta DCI, \Delta DBN$ có:
Chung $\hat D$
$\widehat{DCI}=\widehat{DBN}(=90^o)$
$\to \Delta DCI\sim\Delta DBN(g.g)$
$\to \dfrac{DC}{DB}=\dfrac{DI}{DN}$
$\to \dfrac{DC}{DI}=\dfrac{DB}{DN}$
Mà $\widehat{BDC}=\widehat{IDN}$
$\to \Delta DCB\sim\Delta DIN(c.g.c)$
$\to \widehat{DBC}=\widehat{DNI}$
$\to \widehat{INC}=\widehat{ABC}$
Mà $\widehat{ICN}=\widehat{ACB}(=90^o)$
$\to \Delta CAB\sim\Delta CIN(g.g)$
$\to \dfrac{AB}{IN}=\dfrac{CB}{CN}$
$\to AB.NC=IN.CB$
c.Tương tự câu b chứng minh được $\Delta DIM\sim\Delta DCA(c.g.c)$
$\to \widehat{DMI}=\widehat{DAC}$
$\to 180^o-\widehat{DMI}=180^o-\widehat{DAC}$
$\to \widehat{BAC}=\widehat{IMN}$
Mà $\widehat{ABC}=\widehat{INM}$
$\to \Delta CAB\sim\Delta IMN(g.g)$
$\to \widehat{MIN}=\widehat{BCA}=90^o$
d.Từ câu c
$\to \dfrac{S_{IMN}}{S_{ABC}}=(\dfrac{MN}{AB})^2$
Để $S_{IMN}=2S_{ABC}$
$\to (\dfrac{MN}{AB})^2=2$
$\to MN=AB\sqrt{2}$
Kẻ $EN\perp AM, E\in AM$
$\to ABNE$ là hình chữ nhật
$\to NE=AB$
$\to MN=NE\sqrt2$
$\to \Delta MNE$ vuông cân tại $E$
$\to \widehat{MNE}=\widehat{NME}=45^o$
$\to \widehat{BNM}=45^o$
$\to \widehat{AIC}=45^o$
