Đáp án:
$\begin{array}{l}
a)D{E^2} + D{F^2} = {6^2} + {8^2} = 100\\
E{F^2} = 100\\
\Rightarrow D{E^2} + D{F^2} = E{F^2}\\
\Rightarrow \Delta DEF \bot D\left( {theo\,Pytago} \right)\\
\Rightarrow \left\{ {DK = \dfrac{{DE.DF}}{{EF}} = \dfrac{{6.8}}{{10}} = 4,8\left( {cm} \right)} \right.\\
\Rightarrow F{K^2} = D{F^2} - D{K^2}\\
= {8^2} - 4,{8^2} = 40,96\\
\Rightarrow FK = 6,4\left( {cm} \right)\\
Vậy:DK = 4,8cm;FK = 6,4cm\\
b)Theo\,t/c:\\
\dfrac{{MD}}{{DE}} = \dfrac{{MF}}{{EF}}\\
\Rightarrow \dfrac{{MD}}{6} = \dfrac{{MF}}{8} = \dfrac{{MD + MF}}{{6 + 8}} = \dfrac{{10}}{{14}} = \dfrac{5}{7}\\
\Rightarrow \left\{ \begin{array}{l}
MD = \dfrac{{30}}{7}\left( {cm} \right)\\
MF = \dfrac{{40}}{7}\left( {cm} \right)
\end{array} \right.\\
\Rightarrow EM = \sqrt {D{E^2} + M{D^2}} = \dfrac{{6\sqrt {74} }}{7}\left( {cm} \right)\\
c)\\
EF.\tan \widehat {DEM}\\
= EF.\dfrac{{DM}}{{DE}}\\
Do:\dfrac{{DM}}{{DE}} = \dfrac{{MF}}{{EF}}\\
\Rightarrow EF.\dfrac{{DM}}{{DE}} = MF\\
\Rightarrow MF = EF.\tan \widehat {DEM}
\end{array}$
