Giải thích các bước giải:
a.Ta có $CA\perp AB, BD\perp AB$
$\to\widehat{CAO}=\widehat{DBO}=90^o$
Mà $AC.DB=\dfrac{AB^2}{4}=\dfrac{AB}{2}.\dfrac{AB}{2}=OA.OB$ vì $O$ là trung điểm $AB$
$\to\dfrac{AC}{BO}=\dfrac{AO}{BD}$
$\to\Delta CAO\sim\Delta OBD(c.g.c)$
$\to \widehat{COA}=\widehat{ODB}$
$\to \widehat{COA}+\widehat{DOB}=\widehat{ODB}+\widehat{DOB}$
$\to \widehat{COA}+\widehat{DOB}=90^o$
$\to 180^o-(\widehat{COA}+\widehat{DOB})=90^o$
$\to\widehat{COD}=90^o$
$\to \Delta COD$ vuông ở $O$
b.Ta có:
$\widehat{COD}=\widehat{OBD}=90^o$
Lại có:
$\Delta CAO\sim\Delta OBD$ (câu a)
$\to\dfrac{CO}{OD}=\dfrac{AO}{BD}=\dfrac{OB}{BD}$
$\to\dfrac{CO}{OB}=\dfrac{OD}{BD}$
$\to\Delta BOD\sim\Delta OCD(c.g.c)$