a) Xét \(\Delta ADK\) và \(\Delta CNK\)
Có \(\widehat{AKD}=\widehat{CKN}\) (dđ)
\(\widehat{DAK}=\widehat{NCK}\) (slt của AD // BC )
\(\Rightarrow\) \(\Delta ADK\) \(\infty\) \(\Delta CNK\) (g.g)
b) Xét \(\Delta KAM\) và \(\Delta KCD\)
Có \(\widehat{AKM}=\widehat{CKD}\) (dđ)
\(\widehat{MAK}=\widehat{DCK}\) (slt của AB // CD)
\(\Rightarrow\) \(\Delta KAM\) \(\infty\) \(\Delta KCD\) (g.g)
\(\Rightarrow\dfrac{KA}{KC}=\dfrac{KM}{KD}\left(1\right)\)
Vì \(\Delta ADK\) \(\infty\) \(\Delta CNK\) (cmt)
\(\Rightarrow\dfrac{KA}{KC}=\dfrac{KD}{KN}\left(2\right)\)
(1)(2) \(\Rightarrow\dfrac{KM}{KD}=\dfrac{KD}{KN}\)
\(\Rightarrow KM\cdot KN=KD^2\)
c) Xét \(\Delta DAM\) và \(\Delta NBM\)
Có \(\widehat{DMA}=\widehat{NMB}\) (dđ)
\(\widehat{DAM}=\widehat{NBM}\left(=\widehat{BCD}\right)\)
\(\Rightarrow\) \(\Delta DAM\) \(\infty\) \(\Delta NBM\) (G.G)
\(\Rightarrow\dfrac{AD}{NB}=\dfrac{AM}{BM}\)
.\(\Rightarrow\) \(\dfrac{9}{NB}=\dfrac{6}{4}\)\(\Rightarrow NB=\dfrac{9\cdot4}{6}=6\left(cm\right)\)
Có NB + BC 
CN
\(\Rightarrow\) 6 + 9 = CN \(\Rightarrow\) CN = 15 (cm)
Vì \(\Delta KAM\) \(\infty\) \(\Delta KCD\) (cmt)
\(\Rightarrow\dfrac{S_{\Delta KAM}}{S_{\Delta KCD}}=\left(\dfrac{AM}{CD}\right)^2=\left(\dfrac{6}{10}\right)^2=\dfrac{36}{100}\)
