Giải thích các bước giải:

1.Ta có: $\Delta ABC$ vuông tại $A\to \widehat{ABC}=90^o-\hat C=30^o$

$\to \hat B<\hat C$

$\to AC<AB$

2.Xét $\Delta ABH,\Delta EBH$ có:

Chung $BH$

$\widehat{BHA}=\widehat{BHE}(=90^o)$

$HA=HE$

$\to \Delta ABH=\Delta EBH(c.g.c)$

$\to BA=BE, \widehat{EBH}=\widehat{ABH}=\widehat{ABC}=30^o\to\widehat{ABE}=\widehat{ABH}+\widehat{HBE}=60^o$

$\to \Delta ABE$ đều

3.Xét $\Delta ABC,\Delta EBC$ có:
Chung $BC$

$\widehat{ABC}=\widehat{ABH}=\widehat{EBH}=\widehat{EBC}$

$BA=BE$

$\to\Delta ABC=\Delta EBC(c.g.c)$

$\to\widehat{CEB}=\widehat{CAB}=90^o$

$\to CE\perp EB$

4.Xét $\Delta MAH,\Delta MEH$ có:

Chung $MH$

$\widehat{MHA}=\widehat{MHE}(=90^o)$

$HA=HE$

$\to \Delta MHA=\Delta MHE(c.g.c)$

$\to MA=ME$

Mà $MA=MD\to MD=ME\to \Delta MAE,\Delta MDE$ cân tại $M$

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

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

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

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

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

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

$\to AE\perp DE$

Mà $AH\perp BC\to AE\perp BC$

$\to ED//CB$