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 $$PQ\parallel BC$$ enter image description here

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


Let $R$ and $S$ be the points where $\overleftrightarrow{PQ}$ meets $\overline{CA}$ and $\overline{AB}$, respectively.

enter image description here

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 $\overline{BC}$. $\square$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.