a) Vì P nằm giữa cung AB
$⇒\widehat{ADP}=\widehat{PCB}$
$⇒\widehat{IDK}=\widehat{ADP}=\widehat{PCB}=\widehat{ICK}$
$⇒CDKI$ nội tiếp $⇒\widehat{CID}=\widehat{CKD}$
b) Ta có:
$\widehat{PEF}=\widehat{EPB}+\widehat{PBE}=\widehat{CPB}+\widehat{PBA}=\widehat{CDB}+\widehat{PDB}=\widehat{FDC}$
⇒ CDFE nội tiếp
c) Vì CDKI nội tiếp
$⇒\widehat{DIK}=\widehat{DCK}=\widehat{BAD}⇒IK//AB$