Giải thích các bước giải:

Bài 6:

a.Xét $\Delta ABE,\Delta FBE$ có:

Chung $BE$

$\widehat{ABE}=\widehat{FBE}$

$BA=BE$

$\to \Delta ABE=\Delta FBE(c.g.c)$

$\to \widehat{BFE}=\widehat{BAE}=90^o$

$\to EF\perp BC$

b.Từ a $\to EA=EF$

Xét $\Delta EFC,\Delta EAM$ có:

$EF=EA$

$\widehat{FEC}=\widehat{AEM}$

$EC=EM$

$\to \Delta EFC=\Delta EAM(c.g.c)$

$\to \widehat{MAE}=\widehat{EFC}=90^o$

$\to AM\perp AC$

Mà $AB\perp AC$

$\to B, A, M$ thẳng hàng 

Bài 7:

a.Xét $\Delta AKB,\Delta AKC$ có:

Chung $AK$

$AB=AC$

$KB=KC$

$\to \Delta AKB=\Delta AKC(c.c.c)$

b.Xét $\Delta AKC,\Delta KBH$ có:

$KA=KH$

$\widehat{AKC}=\widehat{BKH}$

$KC=KB$

$\to \Delta AKC=\Delta HKB(c.g.c)$

$\to AC=BH$

c.Từ b $\to \widehat{KAC}=\widehat{KHB}\to AC//HB$

Mà $KM\perp AC, KN\perp HB$

$\to KM\perp AC, KN\perp AC$

$\to M, K, N$ thẳng hàng