$1) ΔBOD$ vuông tại $D$
$⇒\widehat{BOD}+\widehat{BDO}=90°$
Ta có : $\widehat{BOD}+\widehat{COD}+\widehat{AOC}=180°$
Mà %\widehat{COD}=90°$
$⇒\widehat{BOD}+\widehat{AOC}=90°$
$⇒\widehat{BDO}=\widehat{AOC}$
Xét hai tam giác : $ΔACO$ và $ΔBOD$ có :
$\widehat{AOC}=\widehat{BDO}$
$\widehat{CAO}=\widehat{DBO}$
$⇒ΔACO~ΔBDO (g.g)$
$b)$ Theo câu $a) : ΔACO~ΔBOD$
$⇒\dfrac{AC}{OB}=\dfrac{OA}{BD}$
$⇔AC.BD=OA.OB=\dfrac{AB}{2}.\dfrac{AB}{2}=a.a=a^2$
