Giải thích các bước giải:

a.Xét $\Delta ABO,\Delta DOC$ có:

$\widehat{ABO}=\widehat{ABD}=\widehat{ACD}=\widehat{OCD}$

$\widehat{AOB}=\widehat{COD}$

$\to \Delta AOB\sim\Delta DOC(g.g)$

b.Từ a $\to \dfrac{OA}{OD}=\dfrac{OB}{OC}$

Mà $\widehat{AOD}=\widehat{BOC}$(đối đỉnh)

$\to \Delta AOD\sim BOC(c.g.c)$

c.Từ b $\to\widehat{ODA}=\widehat{OCB}$

$\to \widehat{EDB}=\widehat{ECA}$

Xét $\Delta EAC,\Delta EBD$ có:

Chung $\hat E$

$\widehat{EDB}=\widehat{ECA}$

$\to \Delta EDB\sim\Delta ECA(g.g)$

$\to \dfrac{ED}{EC}=\dfrac{EB}{EA}$

$\to EA.ED=EB.EC$

d.Xét $\Delta EAB,\Delta ECD$ có:

Chung $\hat E$

$ \dfrac{ED}{EC}=\dfrac{EB}{EA}$

$\to \Delta EAB\sim\Delta ECD(c.g.c)$