Giải thích các bước giải:

a.Vì $ABCD$ là hình vuông $\to AB=BC=CD=DA, AC\cap BD=O$ là trung điểm mỗi đường và $OA=OB=OC=OD$

Mặt khác $\widehat{CAB}=45^o,\widehat{DBC}=45^o$

$\to \widehat{CAB}=\widehat{DBC}\to \widehat{OAE}=\widehat{OBM}$

Xét $\Delta AOE,\Delta BOM$ có:

$OA=OB$

$\widehat{OAE}=\widehat{OBM}$

$AE=AB-BE=CB-CM=BM$

$\to \Delta AOE=\Delta BOM(c.g.c)$

$\to OE=OM, \widehat{AOE}=\widehat{BOM}$

$\to \widehat{EOM}=\widehat{EOB}+\widehat{BOM}=\widehat{EOB}+\widehat{AOE}=\widehat{AOB}=90^o$ vì $ABCD$ là hình vuông

Do $OE=OM,\widehat{EOM}=90^o\to \Delta OEM$ vuông cân tại $O$

b.Ta có: $AB//CD$

$\to \dfrac{AM}{MN}=\dfrac{MB}{MC}$

Mà $MB=AE, CM=BE$

$\to \dfrac{AM}{MN}=\dfrac{AE}{EB}$

$\to ME//BN$

c.Xét $\Delta BCN,\Delta BCH$ có:

Chung $\hat B$

$\widehat{BHC}=\widehat{BCN}(=90^o)$

$\to \Delta BCH\sim\Delta BNC(g.g)$

$\to \dfrac{BC}{BN}=\dfrac{BH}{BC}$

$\to BC^2=BH\cdot BN$

Xét $\Delta BOC,\Delta BCD$ có:

Chung $\hat B$

$\widehat{BOC}=\widehat{BCD}(=90^o)$

$\to \Delta BOC\sim\Delta BCD(g.g)$

$\to \dfrac{BO}{BC}=\dfrac{BC}{BD}$

$\to BC^2=BO\cdot BD$

$\to BO\cdot BD=BH\cdot BN$

$\to \dfrac{BO}{BN}=\dfrac{BH}{BD}$

Mà $\widehat{OBH}=\widehat{DBN}$

$\to \Delta BOH\sim\Delta BND(c.g.c)$

$\to \widehat{BOH}=\widehat{BND}$

Vì $AB//CD, ME//BN$

$\to \widehat{BEM}=\widehat{BND}$

$\to \widehat{BOH}=\widehat{BEM}$

Gọi $OM\cap BE=F$

Xét $\Delta FBM,\Delta FOE$ có:

Chung $\hat F$

$\widehat{FBM}=\widehat{FOE}(=90^o)$

$\to \Delta FBM\sim\Delta FOE(g.g)$

$\to \dfrac{FB}{FO}=\dfrac{FM}{FE}$

$\to \dfrac{FB}{FM}=\dfrac{FO}{FE}$

Mà $\widehat{EFM}=\widehat{OFB}$

$\to \Delta FEM\sim\Delta FOB(c.g.c)$

$\to \widehat{FEM}=\widehat{FOB}$

$\to \widehat{BEM}=\widehat{BOM}$

$\to \widehat{BOM}=\widehat{BOH}(=\widehat{BEM})$

$\to O, M, H$ thẳng hàng