Giải thích các bước giải:

a.Xét $\Delta AEB,\Delta AFC$ có:

Chung $\hat A$

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

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

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

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

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

c.Vì $BICK$ là hình bình hành

$\to IB//CK, CI//KB$

Do $IB\perp AC, CI\perp AB$

$\to KB\perp AB, KC\perp AC$

Gọi $O$ là trung điểm $AK, AK\cap BE=D$

$\to OB=OA=OK=\dfrac12AK$

      $OC=OA=OK=\dfrac12AK$

$\to OA=OB=OC=OK$

$\to \widehat{AKC}=\widehat{OKC}=\widehat{OCK}=90^o-\widehat{OCA}=90^o-\dfrac12(\widehat{OAC}+\widehat{OCA})=90^o-\dfrac12(\hat A-\widehat{OAB}+\hat C-\widehat{OCB})=90^o-\dfrac12(\hat A+\hat C-\widehat{OAB}-\widehat{OCB})=90^o-\dfrac12(\hat A+\hat C-\widehat{OBA}-\widehat{OBC})=90^o-\dfrac12(\hat A+\hat C-\hat B)=90^o-\dfrac12(180^o-2\hat B)=\hat B=90^o-\widehat{IAF}=\widehat{AIF}$

$\to \Delta AKC\sim\Delta AIF(g.g)$

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

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

Mà $\widehat{AFM}=\widehat{AFE}=\widehat{ACB}=\widehat{ACN}$

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

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

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

$\to MN//IK$