Giải thích các bước giải:

a.Xét $\Delta AEB,\Delta AFC$ có:

Chung $\hat A$

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

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

b.Từ a $\to \dfrac{AE}{AF}=\dfrac{AB}{AC}$

Mà $\widehat{EAF}=\widehat{BAC}$

$\to \Delta AEF\sim\Delta ABC(c.g.c)$

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

c.Vì $BICK$ là hình bình hành $\to IB=CK, CI=KB, BI//CK, CI//KB\to KC\perp AC, KB\perp AB$

$\to \widehat{ABK}=\widehat{ACK}=90^o$

$\to \widehat{ACK}=\widehat{AFI}$

Xét $\Delta BIF, \Delta AFC$ có:

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

$\widehat{BIF}=90^o-\widehat{FBI}=90^o-\widehat{ABE}=\widehat{BAE}=\widehat{FAC}$

$\to \Delta FIB\sim\Delta FAC(g.g)$

$\to\dfrac{BI}{AC}=\dfrac{FI}{AF}$

$\dfrac{CK}{AC}=\dfrac{FI}{AF}$

$\to \Delta FAI\sim\Delta CAK(c.g.c)$

$\to \dfrac{FI}{FA}=\dfrac{CK}{CA}, \dfrac{AI}{AK}=\dfrac{AF}{AC}$

Mặt khác $\widehat{IAF}=\widehat{KAC}\to \widehat{MAF}=\widehat{NAC}$

Vì $\widehat{AFE}=\widehat{ACB}\to \widehat{AFM}=\widehat{ACN}$

$\to \Delta AFM\sim\Delta ACN(g.g)$

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

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

$\to MN//IK$