Giải thích các bước giải:

a.Xét $\Delta ABE, \Delta ACD$ có:

$BE=BC+CE=BC+BD=CD$

$\widehat{ABE}=\widehat{ACD}$ vì $\Delta ABC$ cân tại $A$

$AB=AC$

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

b.Xét $\Delta BDI,\Delta CEK$ có:

$\widehat{DIB}=\widehat{CKE}=90^o$

$BD=CE$

$\widehat{DBI}=\widehat{ABC}=\widehat{ACB}=\widehat{ECK}$

$\to\Delta IBD=\Delta KCE$(cạnh huyền-góc nhọn)

$\to DI=EK, BI=CK\to AI=AB+BI=AC+CK=AK$

$\to IH^2=AH^2-AI^2=AH^2-AK^2=HK^2$

$\to HI=HK$

$\to HD=HI+ID=HK+KE=HE$

c.Xét $\Delta ABK,\Delta ACI$ có:

$AB=AC$

Chung $\hat A$

$AK=AI$

$\to\Delta ABK=\Delta ACI(c.g.c)$

$\to \widehat{AKB}=\widehat{AIC},\widehat{ABK}=\widehat{ACI}$

$\to \widehat{OIB}=\widehat{OKC},\widehat{IBO}=180^o-\widehat{ABK}=180^o-\widehat{ACI}=\widehat{OCK}$

Xét $\Delta OBI,\Delta OCK$ có:
$\widehat{OBI}=\widehat{OCK}$

$BI=CK$

$\widehat{OIB}=\widehat{OKC}$

$\to\Delta OBI=\Delta OCK(g.c.g)$

$\to OB=OC$

$\to\Delta OBC$ cân tại $O$

$\to\widehat{OBC}=\widehat{OCB}$

$\to 180^o-\widehat{OBC}=180^o-\widehat{OCB}$

$\to \widehat{DBO}=\widehat{OCE}$

Xét $\Delta OBD,\Delta OCE$ có:
$OB=OC$

$\widehat{DBO}=\widehat{OCE}$

$BD=CE$

$\to \Delta OBD=\Delta OCE(c.g.c)$

$\to OD=OE\to \Delta oDE$ cân tại $O$

d.Xét $\Delta ABO,\Delta ACO$ có:

Chung $AO$

$AB=AC$

$OB=OC$

$\to \Delta ABO=\Delta ACO(c.c.c)$

$\to \widehat{BAO}=\widehat{CAO}$

$\to AO$ là phân giác $\widehat{BAC}$

e.Ta có $AD=AE, OD=OE, HD=HE\to A,O,H\in$ trung trực của $DE$

$\to A,O,H$ thẳng hàng