a,
Do $ΔABC$ vuông tại $A$
$⇒\widehat{ABC}+\widehat{ACB}=90^o$
$⇒\widehat{ABC}-\dfrac{1}{3}.\widehat{ABC}+\widehat{ACB}-\dfrac{1}{3}.\widehat{ACB}=90^o-\dfrac{1}{3}.(\widehat{ABC}+\widehat{ACB})$
$⇒\widehat{ABC}-\widehat{ABE}+\widehat{ACB}-\widehat{ACD}=90^o-\dfrac{1}{3}.90^o=60^o$
Hay $\widehat{OBC}+\widehat{OCB}=60^o$
Xét $ΔOBC$: $\widehat{OBC}+\widehat{OCB}+\widehat{BOC}=180^o$
$⇒60^o+\widehat{BOC}=180^o$
$⇒\widehat{BOC}=120^o$
$⇒\widehat{DOE}=120^o$ (2 góc đối đỉnh)
Mà $\widehat{BOC}+\widehat{DOE}+\widehat{DOB}+\widehat{EOC}=360^o$
Mà $\widehat{BOC}+\widehat{DOE}=120^o+120^o=240^o$
$\widehat{DOB}=\widehat{EOC}$ (2 góc đối đỉnh)
$⇒240^o+2.\widehat{DOB}=360^o$
$⇒2.\widehat{DOB}=120^o$
$⇒\widehat{DOB}=60^o$
$OK$ là đường phân giác $\widehat{BOC}$
$⇒\widehat{KOB}=\widehat{KOC}=\dfrac{\widehat{BOC}}{2}=\dfrac{120^o}{2}=60^o$
$⇒\widehat{KOB}=\widehat{DOB}=60^o$
Mặt khác $BK$ là đường phân giác $\widehat{OBC}$
$⇒\widehat{KBO}=\widehat{KBC}$
Mà $\widehat{ABC}-\dfrac{1}{3}.\widehat{ABC}=\widehat{OBC}=\widehat{KBO}+\widehat{KBC}$
$⇒2.\widehat{KBO}=\dfrac{2}{3}\widehat{ABC}$
$⇒\widehat{KBO}=\dfrac{1}{3}\widehat{ABC}$
$⇒\widehat{DBO}=\widehat{KOB}$
Xét $ΔDOB$ và $ΔKOB$ có:
$\widehat{DOB}=\widehat{KOB}=60^o$
$OB$ chung
$\widehat{DBO}=\widehat{KOB}$
$⇒ΔDOB=ΔKOB(g.c.g)$
$⇒DO=OK(1)$
Chứng minh tương tự như trên ta được:
$ΔCEO=ΔCKO$
$⇒OE=OK(2)$
Từ $(1)(2)⇒DO=OK=OE$
b,Ta có: $\widehat{DOK}=\widehat{DOB}+\widehat{KOB}=60^o+60^o=120^o$
Ta lại có: $DO=OK⇒ΔDOK$ cân tại $O$
$⇒\widehat{ODK}=\widehat{OKD}=\dfrac{180^o-\widehat{DOK}}{2}=\dfrac{180^o-120^o}{2}=30^o$
Chứng minh tương tự như trên ta có: $\widehat{DOE}=30^o$;$\widehat{DEO}=30^o$;$\widehat{OEK}=30^o$
$⇒\widehat{DOE}+\widehat{ODK}=30^o+30^o=60^o$
Hay $\widehat{KOE}=60^o$
Và $\widehat{DEK}=\widehat{DEO}+\widehat{OEK}=30^o+30^o=60^o$
$⇒\widehat{KOE}=\widehat{DEK}=60^o$
$⇒ΔDEK$ đều
