Consider $\Delta ADF$ and $\Delta AEC$. Since $AD = DE$ and $DF // EC$, then we have
$\begin{array}{rcll}
AF & = & FC & \text{(intercept theorem)} \\
AF & = & \dfrac{1}{2} AC
\end{array}$
Since $AD = DE$ and $AF = FC$, we have
$\begin{array}{rcll}
DF & = & \dfrac{1}{2} EC & \text{(mid-point theorem)} \\
DF & = & \dfrac{1}{2} \times 60 \\
DF & = & 30 \text{ cm}
\end{array}$
Since $\Delta ABC$ is an isosceles triangle with $AB = AC$, we have
$\begin{array}{rcl}
AF & = & \dfrac{1}{2} AB
\end{array}$
In $\Delta ADF$ and $\Delta EDF$,
$\begin{array}{rcll}
AD & = & ED & \text{(given)} \\
DF & = & DF & \text{(common side)} \\
\angle ADF & = & \angle EDF & \text{(adj. $\angle$s on a st. line)}
\end{array}$
Therefore, $\Delta ADF \cong \Delta EDF \ \text{(S.A.S.)}$. Hence, we have
$\begin{array}{rcll}
EF & = & AF & \text{(corr. sides, $\cong \Delta$s)} \\
EF & = & \dfrac{1}{2} AB
\end{array}$
By applying the Pythagoras Theorem to $\Delta DEF$, we have
$\begin{array}{rcl}
EF^2 & = & DF^2 + DE^2 \\
(\dfrac{1}{2} AB)^2 & = & 30^2 + (\dfrac{2}{5} AB)^2 \\
\dfrac{1}{4} AB^2 & = & 900 + \dfrac{4}{25} AB^2 \\
\dfrac{9}{100} AB^ 2 & = & 900 \\
AB^2 & = & 10000 \\
AB & = & 100 \text{ cm}
\end{array}$
Therefore, we have
$\begin{array}{rcl}
EF & = & \dfrac{1}{2} AB \\
EF & = & \dfrac{1}{2} \times 100 \\
EF & = & 50\text{ cm}
\end{array}$
