a) Vì $AB//CD $
$⇒ ΔAOB$ ~ $ΔCOD$
$⇒ \dfrac{OA}{OC}=\dfrac{OB}{OD}$
$⇒ OA.OD=OB.OC$
Xét ΔOAB và ΔOCD có:
$\widehat{OAB}=\widehat{OCD}$
$\widehat{OBA}=\widehat{ODC}$
$⇒ ΔAOB$ ᔕ $ΔCOD$
b) $ΔOAH$ và $ΔCOK$ có:
$\widehat{AHO}=\widehat{CKO}=90^0$
$\widehat{HOA}=\widehat{KOC}$
$⇔$ $\begin{cases}ΔAOH ~ ΔCOK⇒\dfrac{OH}{OK}=\dfrac{OA}{OC}\\\dfrac{OA}{OC}=\dfrac{AB}{CD}\end{cases}$
$⇒ \dfrac{OH}{OK}=\dfrac{AB}{CD}$