Giải thích các bước giải:

a.Xét $\Delta ABE,\Delta ACF$ có:

Chung $\hat A$

$\hat E=\hat F(=90^o)$

$\to \Delta ABE\sim\Delta ACF(g.g)$

b.Ta có: $BK//IC, CK//IB\to BICK$ là hình bình hành

$\to IB=KC$

Xét $\Delta AFC,\Delta BFI$ có:

$\widehat{AFC}=\widehat{BFI}(=90^o)$

$\widehat{FCA}=90^o-\hat A=\widehat{FBI}$

$\to \Delta AFC\sim\Delta IFB(g.g)$

$\to \dfrac{AF}{IF}=\dfrac{AC}{IB}=\dfrac{AC}{KC}$

Mà $CK//EB, BE\perp AC\to \widehat{ACK}=\widehat{AFI}=90^o$

$\to \Delta ACK\sim\Delta AFI(c.g.c)$

$\to \dfrac{FI}{FA}=\dfrac{CK}{CA}$

c.Từ b $\to \dfrac{AI}{AK}=\dfrac{AF}{AC}$

Từ a $\to \dfrac{AE}{AB}=\dfrac{AF}{AC}\to \Delta AEF\sim\Delta ACB(c.g.c)$

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

$\to \widehat{AFM}=\widehat{ACN}$

Từ b $\to \widehat{IAF}=\widehat{KAC}\to \widehat{MAF}=\widehat{NAC}$

$\to \Delta AMF\sim\Delta ANC(g.g)$

$\to\dfrac{AM}{AN}=\dfrac{AF}{AC}=\dfrac{AI}{AK}$

$\to \dfrac{AM}{AI}=\dfrac{AN}{AK}$

$\to MN//IK$