Giải thích các bước giải:
a.Ta có :
$\Diamond MDCA, DNBC$ nội tiếp
$\to \widehat{MCD}+\widehat{DCN}=\widehat{MAD}+\widehat{DBN}=\widehat{DBA}+\widehat{DBN}=90^o$
$\to \Diamond DPCQ$ nội tiếp
$\to \widehat{DPQ}=\widehat{DCQ}=\widehat{DBN}=\widehat{DAB}\to PQ//AB$
b.Ta có :
$\dfrac{CA}{CM}=\cos \widehat{MCA}$
$\dfrac{CB}{CN}=\cos \widehat{NCB}=\sin \widehat{MCA},(\widehat{MCA}+\widehat{NCA}=90^o)$
$\to \dfrac{CA}{CM}.\dfrac{CB}{CN}=\cos \widehat{MCA}.\sin \widehat{MCA}\le \dfrac{1}2(\cos^2 \widehat{MCA}+\sin^2 \widehat{MCA})=\dfrac{1}2$
$\to CM.CN\ge 2CA.CB$