Giải thích các bước giải:

a.Ta có: $\widehat{xOy},\widehat{yOz}$ kề và bằng nhau

$\to Oy$ là phân giác $\widehat{xOz}$

Mà $Om, On$ là phân giác $\widehat{xOy},\widehat{yOz}$

$\to \widehat{xOm}=\widehat{mOy}=\dfrac12\widehat{xOy}=\dfrac12\widehat{yOz}=\widehat{yOn}=\widehat{nOz}$ 

Xét $\Delta OAB,\Delta OBC$ có:

Chung $OB$

$\widehat{BOA}=\widehat{mOx}=\widehat{mOy}=\widehat{BOC}$

$OA=OC$

$\to \Delta OAB=\Delta OCB(c.g.c)$

$\to AB=BC$

Tương tự chứng minh được $BC=CD, DC=DE$

$\to AB=BC=CD=DE$

b.Xét $\Delta OCA,\Delta OCE$ có:

Chung $OC$

$\widehat{AOC}=\widehat{COE}$

$OA=OE$

$\to \Delta OAC=\Delta OEC(c.g.c)$

$\to AC=CE$

Mà $BA=CD, BC=DE$

$\to \Delta BAC=\Delta DCE$

$\to \widehat{BAC}=\widehat{DCE}$

c.Xét $\Delta AOD,\Delta BOE$ có:

$OA=OB$

$\widehat{AOD}=\widehat{AOB}+\widehat{BOD}=\widehat{DOE}+\widehat{BOC}=\widehat{BOE}$

$OD=OE$

$\to \Delta OAD=\Delta OBE(c.g.c)$

$\to AD=BE$

d.Ta có: $OC=OE, DC=DE$

$\to O, D\in$ trung trực $CE$

$\to OD$ là trung trực $CE$

$\to OD\perp CE$

Tương tự chứng minh được $OB$ là trung trực $AC\to OB\perp AC$

e.Ta có:

$\widehat{ABC}=2\widehat{ABO}$

$\widehat{BAD}=\widehat{BAO}-\widehat{DAO}=\widehat{ABO}-(90^o-\dfrac12\widehat{AOD})$

$\to\widehat{BAD}=\widehat{BAO}-(90^o-\dfrac12\cdot 3\widehat{AOB})$

$\to \widehat{BAD}=\widehat{BAO}-90^o+\dfrac32\widehat{AOB}$

$\to \widehat{ABC}+\widehat{BAD}=2\widehat{ABO}+\widehat{BAO}-90^o+\dfrac32\widehat{AOB}$

$\to \widehat{ABC}+\widehat{BAD}=3\widehat{ABO}-90^o+\dfrac32\widehat{AOB}$

$\to \widehat{ABC}+\widehat{BAD}=3\widehat{ABO}+\dfrac32\widehat{AOB}-90^o$

$\to \widehat{ABC}+\widehat{BAD}=3(\widehat{ABO}+\dfrac12\widehat{AOB})-90^o$

$\to \widehat{ABC}+\widehat{BAD}=3\cdot 90^o-90^o$ vì $\Delta OAB$ cân tại $O$

$\to \widehat{ABC}+\widehat{BAD}=180^o$

$\to BC//AD$

Tương tự chứng minh được $EB//CD$

g.Ta có: $OA=OE\to \Delta OAE$ cân tại $O$

                $OC$ là phân giác $\widehat{AOE}$

$\to OC$ là trung trực $AE$

Xét $\Delta IBA,\Delta IDE$ có:

$\widehat{BAI}=180^o-\widehat{ABC}=180^o-2\widehat{ABO}=180^o-2\widehat{ODE}=180^o-\widehat{CDE}=\widehat{DEI}$

$AB=DE$

$\widehat{ABI}=180^o-\widehat{BAI}-\widehat{BIA}=180^o-\widehat{IED}-\widehat{DIE}=\widehat{IDE}$

$\to \Delta IAB=\Delta IED(g.c.g)$

$\to IA=IE$

$\to I\in$ trung trwujc $AE$

$\to I\in OC$

$\to I\in Oy$