Giải thích các bước giải:

a.Xét $\Delta ABI,\Delta AMI$ có:

Chung $AI$

$\widehat{IAB}=\widehat{IAM}$

$AB=AM$

$\to \Delta AIB=\Delta AIM(c.g.c)$

$\to \widehat{AIB}=\widehat{AIM}$

Mà $\widehat{AIB}+\widehat{AIM}=180^o$

$\to \widehat{AIB}=\widehat{AIM}=90^o$

$\to AI\perp BM$

b.Ta có: $AD=2AB=AC$

$\to \Delta ADC$ cân tại $A$

Vì $AM=AB\to \Delta ABM$ cân tại $A$

$\to \widehat{ABM}=90^o-\dfrac12\hat A=\hat D$

$\to BM//DC$

c.Xét $\Delta ABK,\Delta AMK$ có:

Chung $AK$

$\widehat{KAB}=\widehat{KAM}$

$AB=AM$

$\to \Delta AKB=\Delta AKM(c.g.c)$

$\to KB=KM, \widehat{ABK}=\widehat{AMK}$

$\to \widehat{KBD}=180^o-\widehat{ABK}=180^o-\widehat{AMK}=\widehat{KMC}$

Xét $\Delta KBD,\Delta KMC$ có:

$KB=KM$

$\widehat{KBD}=\widehat{KMC}$

$BD=CM$

$\to \Delta KBD=\Delta KMC(c.g.c)$

$\to \widehat{BKD}=\widehat{MKC}$

$\to D, K, M$ thẳng hàng