How prove this $PQ\parallel BC$ Question:
if the point $P$ in  $\triangle ABC$,such $PB\cap AC=E, PC\cap AB=F$, and $PK\parallel AB, PL\parallel AC$, and
$L, F\in AB, K, E\in AC, EF\cap KL=Q$, show that
$$PQ\parallel BC$$

My idea: $$\Longleftrightarrow \dfrac{EQ}{ED}=\dfrac{EP}{EB}$$
since 
$$\dfrac{EP}{EB}=\dfrac{EK}{EA}$$
so
$$\Longleftrightarrow\dfrac{EQ}{ED}=\dfrac{EK}{EA}$$
$$\Longleftrightarrow LQ\parallel AD$$
then I can't prove $LQ\parallel AD$, someone can help? Thank you
or can  see:http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=574067
 A: Let $R$ and $S$ be the points where $\overleftrightarrow{PQ}$ meets $\overline{CA}$ and $\overline{AB}$, respectively.

Lines $\overleftrightarrow{RS}$, $\overleftrightarrow{EF}$, $\overleftrightarrow{KL}$, $\overleftrightarrow{QA}$ concur at $Q$. Therefore, the cross ratios $(A, K; E, R)$ and $(A, L; F, S)$ agree.
$$\frac{|\overline{AE}|\;|\overline{KR}|}{|\overline{KE}|\;|\overline{AR}|} = \frac{|\overline{AF}|\;|\overline{LS}|}{|\overline{LF}|\;|\overline{AS}|} \qquad (\star)$$
Since $\overline{PK}\parallel\overline{AB}$, we have $\triangle AEB \sim \triangle KEP$ and $\triangle KRP \sim \triangle ARS$, so that
$$\frac{|\overline{AE}|}{|\overline{KE}|} = \frac{|\overline{AB}|}{|\overline{KP}|} \qquad \text{and} \qquad \frac{|\overline{KR}|}{|\overline{AR}|} = \frac{|\overline{KP}|}{|\overline{AS}|}$$
whence the left-hand side of $(\star)$ becomes
$$\frac{|\overline{AB}|}{|\overline{KP}|} \frac{|\overline{KP}|}{|\overline{AS}|} = \frac{|\overline{AB}|}{|\overline{AS}|}$$
Likewise for the right-hand side of $(\star)$, so that $(\star)$ itself becomes
$$\frac{|\overline{AB}|}{|\overline{AS}|} = \frac{|\overline{AC}|}{|\overline{AR}|} \qquad (\star\star)$$
By the Side-Angle-Side Similarity Theorem, this says that $\triangle ABC \sim \triangle ASR$, and we may conclude that $\overline{RS}$ ---and thus $\overline{PQ}$--- is parallel to $\overline{BC}$. $\square$
