a) Ta có: $\widehat{DAO} = \widehat{ODC}$ (cùng phụ $\widehat{ODA}$)
$\widehat{AOD} = \widehat{COD} = 90^o$
$\Rightarrow ΔDAO = ΔCDO \, (g.g)$
$\Rightarrow \dfrac{OA}{OD} = \dfrac{OD}{OC}$
$\Rightarrow OD^2 = OA.OC$
$\Rightarrow OD = \sqrt{OA.OC} = \sqrt{45.125.} = 75$
Ta có: $\widehat{DAO} = \widehat{OBA}$ (cùng phụ $\widehat{OAB}$)
$\widehat{DOA} = \widehat{AOB} = 90^o$
$\Rightarrow ΔAOD\sim ΔBOA \, (g.g)$
$\Rightarrow \dfrac{AO}{OB} = \dfrac{OD}{OA}$
$\Rightarrow OB = \dfrac{OA^2}{OD} = \dfrac{45^2}{75} = 27$
$\Rightarrow BD = OD + OB = 75 + 27 = 102$
b) Kẻ $OH\perp CD$
$\Rightarrow OH = d(O;CD)$
$\Rightarrow OH//AD$
Áp dụng định lý Thales, ta được:
$\dfrac{CO}{CA} = \dfrac{OH}{AD}$ $(*)$
Xét $ΔDAO$ và $ΔDBA$ có:
$\widehat{D}:$ góc chung
$\widehat{O} = \widehat{A} = 90^o$
Do đó $ΔDAO \sim ΔDBA\, (g.g)$
$\Rightarrow \dfrac{DA}{DB} = \dfrac{DO}{DA}$
$\Rightarrow DA^2 = DO.BD$
$\Rightarrow DA = \sqrt{DO.DB} = \sqrt{75.102} = 15\sqrt{34}$
$(*) \Rightarrow OH = \dfrac{OC.AD}{AC} = \dfrac{125.15\sqrt{34}}{125 + 45} = \dfrac{375\sqrt{34}}{34}$
