Giải thích các bước giải:
a) Ta có:
Xét $\Delta AKE,\Delta CFE$ có:
$\begin{array}{l}
\left\{ \begin{array}{l}
AE = CE\\
\widehat {AEK} = \widehat {CEF}\left( {dd} \right)\\
EK = EF
\end{array} \right.\\
\Rightarrow \Delta AKE = \Delta CFE\left( {c.g.c} \right)
\end{array}$
$ \Rightarrow AK = CF$
b) Ta có:
$\begin{array}{l}
\Delta AKE = \Delta CFE\left( {c.g.c} \right)\\
\Rightarrow \widehat {KAE} = \widehat {FCE}
\end{array}$
$ \Rightarrow AK//CF$ ($2$ góc so le trong bằng nhau)
c) Ta có:
Xét $\Delta BKC,\Delta FCK$ có:
$\begin{array}{l}
\left\{ \begin{array}{l}
BK = FC\left( { = AK} \right)\\
\widehat {BKC} = \widehat {FCK}\left( {slt} \right)\\
KCchung
\end{array} \right.\\
\Rightarrow \Delta BKC = \Delta FCK\left( {c.g.c} \right)\\
\Rightarrow \widehat {BCK} = \widehat {FKC}\\
\Rightarrow KF//BC
\end{array}$
d) Ta có:
$\begin{array}{l}
\Delta BKC = \Delta FCK\left( {c.g.c} \right)\\
\Rightarrow BC = KF\\
\Rightarrow BC = 2KE\\
\Rightarrow KE = \dfrac{1}{2}BC
\end{array}$
