a.Xét$\triangle ADK và \triangle CNK$ có:
$\widehat{ADK}=\widehat{CNK}(SLT)$
$\widehat{AKD}=\widehat{CKN}$(đối đỉnh).
$\rightarrow \triangle ADK\sim \triangle CNK(g-g)\rightarrow \frac{KA}{KC}=\frac{DK}{KN}\rightarrow KA.KN=KD.KC$
b.Xét $\triangle DAM và\triangle DCN$ có:
$\widehat{DAM}=\widehat{DCN}$
$\widehat{ADM}=\widehat{DNC}$
$\rightarrow \triangle DAM\sim \triangle NCD(g-g)\rightarrow \frac{DA}{NC}=\frac{DM}{DN}\rightarrow DA.DN=NC.DM$.
c.Xét $\triangle AKM và \triangle CKD$ có:
$\widehat{AKM}=\widehat{CKD}$(đối đỉnh)
$\widehat{KAM}=\widehat{KCD}(SLT)$
$\rightarrow \triangle AKM\sim \triangle CKD\rightarrow \frac{AK}{CK}=\frac{KM}{KD}$
mà$\frac{KA}{KC}=\frac{DK}{KN} (cmt)$
$\rightarrow \frac{KM}{KD}=\frac{KD}{KN}\rightarrow KD^{2}=KM.KN.$
d.$AB=10(cm)\rightarrow CD=10(cm)$
Có$\triangle AKM\sim \triangle CKD(cmt)$
$\rightarrow \frac{S_{AKM}}{S_{CKD}}=(\frac{AM}{CD})^{2}=(\frac{6}{10})^{2}=\frac{9}{25}$