Giải thích các bước giải:
a.Xét $\Delta ABM,\Delta ACM$ có:
Chung $AM$
$AB=AC$
$MB=MC$ vì $M$ là trung điểm $BC$
$\to\Delta ABM=\Delta ACM(c.c.c)$
b.Từ câu a
$\to\widehat{AMB}=\widehat{AMC}$
Mà $\widehat{AMB}+\widehat{AMC}=\widehat{BMC}=180^o$
$\to \widehat{AMB}=\widehat{AMC}=90^o$
$\to AM\perp BC$
c.Xét $\Delta AMB,\Delta CMD$ có:
$MB=MC$
$\widehat{AMB}=\widehat{CMD}$
$MA=MD$
$\to\Delta AMB=\Delta DMC(c.g.c)$
$\to \widehat{BAM}=\widehat{MDC}\to\widehat{IAM}=\widehat{DMK}$
Xét $\Delta AIM,\Delta DKM$ có:
$AI=DK$
$\widehat{IAM}=\widehat{DMK}$
$MA=MD$
$\to\Delta AIM=\Delta DKM(c.g.c)$
$\to\widehat{AMI}=\widehat{DMK}$
$\to I,M,K$ thẳng hàng
