Giải thích các bước giải:

a.Xét $\Delta AHB,\Delta AHC$ có:

Chung $AH$

$AB=AC$

$HB=HC$

$\to \Delta AHB=\Delta AHC(c.c.c)$

$\to \widehat{AHB}=\widehat{AHC}$

Mà $\widehat{AHB}+\widehat{AHC}=180^o$

$\to \widehat{AHB}=\widehat{AHC}=90^o$

$\to AH\perp BC$

b.Xét $\Delta ABM, \Delta ACN$ có:

$AB=AC$

$\widehat{ABM}=180^o-\widehat{ABC}=180^o-\widehat{ACB}=\widehat{ACN}$

$AB=AC$

$\to \Delta ABM=\Delta ACN(c.g.c)$

c.Từ b $\to AM=AN, \hat M=\hat N, \widehat{BAM}=\widehat{CAN}\to \widehat{BAI}=\widehat{CAK}$

Xét $\Delta ABI,\Delta ACK$ có:

$\hat I=\hat K(=90^o)$

$\widehat{BAI}=\widehat{CAK}$

$AB=AC$

$\to \Delta ABI=\Delta ACK$(cạnh huyền-góc nhọn)

$\to IA=AK$

$\to \Delta AIK$ cân tại $A$

Vì $MA=AN\to \Delta AMN$ cân tại $A$

$\to \widehat{AIK}=90^o-\dfrac12\widehat{IAK}=90^o-\dfrac12\widehat{MAN}=\widehat{AMN}$

$\to IK//MN$