a) $\widehat{ABD}$ = $\widehat{ACD}$(gt) hay $\widehat{ABO}$ = $\widehat{OCD}$
Xét $\triangle$AOB và $\triangle$DOC có:
$\widehat{ABO}$ = $\widehat{OCD}$ (cmt)
$\widehat{AOB}$ = $\widehat{DOC}$ (đối đỉnh)
Vậy $\triangle$AOB $\backsim$ $\triangle$DOC (g.g)
b) Vì $\triangle$AOB $\backsim$ $\triangle$DOC suy ra
$\dfrac{AO}{OB}$ = $\dfrac{DO}{OC}$
Xét $\triangle$AOD và $\triangle$BOC có:
$\dfrac{AO}{OB}$ = $\dfrac{DO}{OC}$ (cmt)
$\widehat{AOD}$ = $\widehat{BOC}$ (đối đỉnh)
Vậy $\triangle$AOD $\backsim$ $\triangle$BOC (c.g.c)
c) Vì $\triangle$AOD $\backsim$ $\triangle$BOC suy ra
$\widehat{ADO}$ = $\widehat{BCO}$ hay $\widehat{EDB}$ = $\widehat{ECA}$
Xét $\triangle$EDB và $\triangle$ECA có:
$\widehat{E}$ chung
$\widehat{EDB}$ = $\widehat{ECA}$ (cmt)
Vậy $\triangle$EDB $\backsim$ $\triangle$ECA (g.g)
$\Rightarrow$ $\dfrac{ED}{EC}$ = $\dfrac{EB}{EA}$
$\Rightarrow$ ED.EA = EC.EB (đpcm)