Euclidean Geometric Problem: prove that line segment ratio is $1:2$ I'm working on this plane geometry problem:
$ \bigtriangleup AEF$ is a triangle, $\angle A = 60^\circ$, $\angle F = 40^\circ$, $\angle E = 80^\circ$, points $K$ and $C$ are on line $AF$, $\angle CEF=20^\circ$, $\angle KEF=40^\circ$, $O$ is the mid-point of $AK$, prove that the length of $OC$ is half of $EF$.
I can prove this by using trigonometric formulas. My question is: is it possible to prove it without using the knowledge of trigonometric formulas?
Here is a figure I draw using Geogebra.

 A: Let $\overline{EF}=1$ for simplicity. Also note that $\triangle AEC$ is equilateral.
Let $H$ on $EF$ such that $\overline{EH} = \overline{EC}$.

Internal Bisector Theorem on the isosceles triangle $\triangle EKF$ gives
$$\overline{KF} = \frac{\overline{KC}}{\overline{KF}-\overline{KC}}.\tag{1}\label{2}$$
Similarity $\triangle CHF \sim \triangle EKF$ leads to
$$\overline{KF}=\frac{1-\overline{AC}}{\overline{KF}-\overline{KC}}.\tag{2}\label{3}$$
Equating \eqref{2} and \eqref{3} gives
$$\overline{KC} = 1-\overline{AC},\tag{3}\label{4}$$
which is equivalent to your thesis. In fact we have
\begin{eqnarray}
\overline{OC} &=& \frac{\overline{AC}-\overline{KC}}2+\overline{KC}=\\
&=&\frac{\overline{AC}+\overline{KC}}2 =\\
&\stackrel{\eqref{4}}{=}&\frac12.
\end{eqnarray}
A: 
Let $K'$ be a point on $AE$ such that $KK' \perp AE$.
Since $\angle AK'K = 90^\circ$, $AK' = AO$.
As $\triangle AK'O$ and $\triangle AEC$ are equilateral triangles, $OC = K'E$.

Let $K''$ be a point on $EF$ such that $KK'' \perp EF$.
Since $\triangle KEK' \cong \triangle KEK'' \cong \triangle KFK''$,
$OC = K'E = \frac{EF}{2}$
$\therefore$ The length of $OC$ is half of $EF$.
