a) $\widehat{AEB}=90^{0} ,\widehat{ACB}=90^{0}\Rightarrow \widehat{BCF}+\widehat{DEF}=180^{0}\Rightarrow $tứ giác CDEF nội tiếpb)$\triangle ACD\sim \triangle BED(g.g)\Rightarrow \frac{AD}{CD}=\frac{DE}{DB}\Rightarrow DA.DE=DB.DC$
c)CDEF nội tiếp$ \Rightarrow \widehat{CFD}=\widehat{CED}=\widehat{CBA}$
mà $\triangle OBC$cân tại O$\Rightarrow \widehat{ABC}=\widehat{OCB}$
$\Rightarrow$ đpcm