Giải thích các bước giải:
a.Ta có $AB\perp AC, HM//AC, HN//AB$
$\to HM\perp AB, HN\perp AC$
Xét $\Delta AME, \Delta AMH$ có:
Chung $AM$
$\widehat{AME}=\widehat{AMH}(=90^o)$ vì $HM\perp AB$
$ME=MH$ vì $M$ là trung điểm $HE$
$\to \Delta AME=\Delta AMH(c.g.c)$
$\to AE=AH, \widehat{EAM}=\widehat{MAH}$
$\to \widehat{EAH}=\widehat{EAM}+\widehat{MAH}=2\widehat{MAH}=2\widehat{BAH}$
Tương tự chứng minh được $AH=AD,\widehat{HAD}=2\widehat{HAC}$
$\to AE=AD(=AH)$
$\widehat{EAD}=\widehat{EAH}+\widehat{HAD}=2\widehat{BAH}+2\widehat{HAC}=2\widehat{BAC}=2\cdot 90^o=180^o\to E,A,D$ thẳng hàng
$\to A$ là trung điểm $DE$
b.Xét $\Delta AMH,\Delta ANH$ có:
$\widehat{MAH}=\widehat{AHN}$ vì $HN//AB$
Chung $AH$
$\widehat{MHA}=\widehat{HAN}$ vì $HM//AC$
$\to \Delta AHM=\Delta HAN(g.c.g)$
$\to AN=MH, AM=HN$
Xét $\Delta AME, \Delta MAN$ có:
Chung $AM$
$\widehat{EMA}=\widehat{MAN}=90^o$
$ME=AN(=MH)$
$\to \Delta AME=\Delta MAN(c.g.c)$
$\to \widehat{NMA}=\widehat{EAM}\to MN//EA\to MN//DE$
Từ câu a $\to \widehat{EAM}=\widehat{MAH}\to \widehat{EAB}=\widehat{BAH}$
Xét $\Delta AEB,\Delta HAB$ có:
Chung $AB$
$\widehat{EAB}=\widehat{BAH}$
$EA=AH$
$\to\Delta ABE=\Delta ABH(c.g.c)$
$\to \widehat{AEB}=\widehat{AHB}=90^o\to BE\perp DE$
Tương tự chứng minh được $CD\perp DE$
$\to BE//CD$
c.Ta có $HM//AC, HN//AB, AB\perp AC\to HN\perp HM$
$\to DH\perp HE$
