How to prove that $DK=DL$

Question

In $\triangle ABC$ the cevians $AD$,$BE$ and $CF$ are concurrent at the point $O$. $DL\parallel EF$ and $BE$ intersects $DL$ at $K$. How to prove that $DK = DL$.
I applied menelaus theorem twice in the $\triangle KEL$ and got a relation $\frac{DL^2} {DK^2}=\frac{BE} {BK}\cdot \frac{CL} {CE} \cdot \frac{AL} {AE} \cdot \frac{EO} {KO}$ and also tried to use the ratio of similar triangles but no result. any ideas on utilizing the ratio of similar triangles.

Answer 1

Following timon92, we are going to prove $\dfrac{DA}{AJ}=\dfrac{DO}{OJ}$.

Consider the figure (only add point $M$, which is an intersection of lines $EF$ and $BC$), we focus on $\triangle MDJ$, and apply Menelaus's theorem 4 times: \begin{align} \color{blue}{\frac{MB}{BD}}\color{red}{\frac{DA}{AJ}}\frac{JF}{FM}&=1;\tag{1}\\ \color{blue}{\frac{MB}{BD}}\color{orange}{\frac{DO}{OJ}}\color{green}{\frac{JE}{EM}}&=1;\tag{2}\\ \color{purple}{\frac{MC}{CD}}\color{orange}{\frac{DO}{OJ}}\frac{JF}{FM}&=1;\tag{3}\\ \color{purple}{\frac{MC}{CD}}\color{red}{\frac{DA}{AJ}}\color{green}{\frac{JE}{EM}}&=1.\tag{4} \end{align} Combining (1) and (4), and combining (2) and (3), respectively, we obtain \begin{align} \left(\color{red}{\frac{DA}{AJ}}\right)^2 =\left(\color{blue}{\frac{MB}{BD}}\frac{JF}{FM}\color{purple}{\frac{MC}{CD}}\color{green}{\frac{JE}{EM}}\right)^{-1} =\left(\color{orange}{\frac{DO}{OJ}}\right)^2. \end{align}

Answer 2

You may want to start like this: Since $EF \parallel DL$, we have $\dfrac{DL}{JE}=\dfrac{DA}{AJ}$ and $\dfrac{DK}{JE} = \dfrac{DO}{OJ}$. The problem reduces to showing that $\dfrac{DA}{AJ}=\dfrac{DO}{OJ}$.