Giải thích các bước giải:
a.Ta có $OC//O'D\to \widehat{COA}+\widehat{AO'D}=180^o$
$\to\widehat{CAO}+\widehat{DAO'}=90^o-\dfrac12\widehat{COA}+90^o-\dfrac12\widehat{DO'A}$
$\to\widehat{CAO}+\widehat{DAO'}=180^o-\dfrac12(\widehat{COA}+\dfrac12\widehat{DO'A})$
$\to\widehat{CAO}+\widehat{DAO'}=180^o-\dfrac12\cdot 180^o$
$\to\widehat{CAO}+\widehat{DAO'}=90^o$
$\to\widehat{CAD}=180^o-(\widehat{CAO}+\widehat{DAO'})=90^o$
$\to AD\perp AC$
b.Ta có $(O), (O')$ tiếp xúc ngoài tại $A\to OO'=OA+AO'=3+2=5$
Ta có $O'D//OC$
$\to \dfrac{KO'}{KO}=\dfrac{O'D}{OC}=\dfrac23$
$\to \dfrac{KO'}{KO-KO'}=\dfrac{2}{3-2}$
$\to \dfrac{KO'}{OO'}=2$
$\to KO'=2OO'=10$