Using Thevenin theorem, find the equivalent Thevenin circuit for the highlighted circuit external to $R_L$.
Solving for $R_{TH}$
$$\begin{align}
R_{TH} &= (R_1 //R_2) + R_3\\
&= \frac{10\Omega}{2} + 20\Omega \\
&= 25\Omega
\end{align}
$$
Solving for $E_{TH}$
$$\begin{align}
\text{Voltage divider rule:} \rightarrow E_{TH} &= \frac{R_2}{R_1+R_2} \cdot E_1\\
&= \frac{10\Omega}{(10+10)\Omega}\cdot 50V\\
&= \frac{10\Omega}{20\Omega}\cdot 50V\\
&= 25V
\end{align}
$$