Đáp án:
Giải thích các bước giải:
ΔABH∞ΔBCK
⇒\[\frac{{BH}}{{CK}} = \frac{{AH}}{{BK}} = \frac{{BA}}{{CB}}\]
⇒\[\frac{{BH.ta{n^2}\widehat {BAC}}}{{BK}} = \frac{{BH.{{(\frac{{BC}}{{BA}})}^2}}}{{BK}} = \frac{{BH.{{(\frac{{CK}}{{BH}})}^2}}}{{BK}} = \frac{{C{K^2}}}{\begin{gathered}
BH.BK \hfill \\
\hfill \\
\end{gathered} }\]\[ = \frac{{C{K^2}}}{{CK.AH}} = \frac{{CK}}{{AH}}\](1)
Mặt khác: Δ AHM∞ΔCKM
\[\begin{gathered}
\Rightarrow \frac{{CK}}{{AH}} = \frac{{KM}}{{HM}} = \frac{{CM}}{{MA}}(2) \hfill \\
TU(1)(2) \Rightarrow \frac{{BH.tan{}^2\widehat {BAC}}}{{BK}} = \frac{{MC}}{{MA}} \hfill \\
\end{gathered} \]
