\(a,AB=AK+KB=6\left(cm\right)\)
Áp dụng HTL tam giác
\(\left\{{}\begin{matrix}BC^2=BK\cdot AB=24\\AC^2=AK\cdot AB=12\\CK^2=AK\cdot KB=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}BC=2\sqrt{6}\left(cm\right)\\AC=2\sqrt{3}\left(cm\right)\\CK=\sqrt{6}\left(cm\right)\end{matrix}\right.\)
\(b,\) Dễ thấy \(CMKN\) là hcn nên \(\widehat{CNM}=\widehat{NCK}\)
Mà \(\widehat{CNM}+\widehat{CMN}=\widehat{NCK}+\widehat{CBK}\left(=90^0\right)\)
Do đó \(\widehat{CMN}=\widehat{CBK}\)
\(\left\{{}\begin{matrix}\widehat{CMN}=\widehat{CBK}\\\widehat{ACB}.chung\end{matrix}\right.\Rightarrow\Delta CMN\sim\Delta CBA\left(g.g\right)\\ \Rightarrow\dfrac{CM}{CB}=\dfrac{CN}{CA}\Rightarrow CM\cdot CA=CN\cdot CB\)
\(c,\) Áp dụng HTL tam giác
\(\left\{{}\begin{matrix}AC^2\cdot BK=BK\cdot AK\cdot AB\\BC^2\cdot AK=AK\cdot BK\cdot AB\end{matrix}\right.\Rightarrow AC^2\cdot BK=BC^2\cdot AK\)
