Giải thích các bước giải:

a.Ta có: $10^2=8^2+6^2$

$\to BC^2=AB^2+AC^2$

$\to\Delta ABC$ vuông tại $A$

b.Xét $\Delta MAB,\Delta MCD$ có:

$MA=MD$

$\widehat{AMB}=\widehat{CMD}$

$MB=MC$

$\to\Delta MAB=\Delta MDC(c.g.c)$

$\to \widehat{MAB}=\widehat{MDC}$

$\to AB//CD$

Do $\Delta ABC $ vuông tại $A\to AB\perp AC\to AC\perp CD$

c.Xét $\Delta AHC,\Delta EHC$ có:

Chung $CH$

$\widehat{AHC}=\widehat{EHC}(=90^o)$

$HA=HE$

$\to\Delta CAH=\Delta CEH(c.g.c)$

$\to CA=CE$

$\to\Delta CAE$ cân tại $C$

d.Xét $\Delta MAC,\Delta MBD$ có:

$MA=MD$

$\widehat{AMC}=\widehat{BMD}$

$MC=MB$

$\to\Delta MAC=\Delta MDB(c.g.c)$

$\to BD=AC$

$\to BD=CE(=AC)$

e.Từ câu c $\to \widehat{ACH}=\widehat{HCE}$

                 d $\to \widehat{MDB}=\widehat{MAC}\to BD//AC$

$\to \widehat{MBD}=\widehat{MCA}=\widehat{ACH}=\widehat{HCE}=\widehat{MCE}$

Xét $\Delta DMB,\Delta EMC$ có:

$MB=MC$

$\widehat{MBD}=\widehat{MCE}$

$BD=CE$

$\to\Delta DMB=\Delta EMC(c.g.c)$

$\to MD=ME$

Mà $MD=MA\to MD=ME=MA$

$\to \Delta MAE,\Delta MDE$ cân tại $M$

$\to \widehat{MAE}=\widehat{MEA}, \widehat{MED}=\widehat{MDE}$

$\to \widehat{AED}=\widehat{AEM}+\widehat{MED}=\widehat{MAE}+\widehat{MDE}$

$\to \widehat{AED}=\widehat{DAE}+\widehat{ADE}$

$\to 2\widehat{AED}=\widehat{AED}+\widehat{DAE}+\widehat{ADE}$

$\to 2\widehat{AED}=180^o$

$\to \widehat{AED}=90^o$

$\to DE\perp AE$

Mà $AH\perp BC\to AE\perp BC$

$\to BC//DE$