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Fig. To prove :

Fix points $A,C$ and set point $B$ to be the midpoint of segment $AC$. Fix point $Y$ (anywhere) and consider an arbitrary point $X$ on line $YB$. If $P$ and $Q$ are the intersection points of two lines $AY \cap CX$ , $CY \cap AX$ respectively, show that line $PQ$ is parallel to line $AC$.

I first proved converse, to see what geometry this figure has in structure.And that if trapezoid $APQC$ is given and If $Y$ and $X$ are the intersection points of the sides of the trapezoid $AP\cap CQ$ and $AQ\cap PC$ respectively, then line $YX$ passes through midpoints of segment $AC$ and $PQ$. However this didn't help much.

Im trying to solve this by looking for some formula related with a point $B$. So $PQ$ and $AC$ will be parallel when $B$ is midpoint of $AC$.

I took hours with this problem and I couldn't find a clue.

What would be the proof look like?

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  • $\begingroup$ AHA! I foud it. $\endgroup$ – Solvable Potato Jan 15 '19 at 5:09
  • $\begingroup$ I made the title more descriptive and reworded some of your work to make it more legible (and turned the formulas into $\LaTeX$). Please feel free to roll back the edit if it changes your meaning in any way. Also, if you've just solved the problem, feel free to add your solution as an answer! That way others who come across the same question can read your solution. $\endgroup$ – Carl Schildkraut Jan 15 '19 at 5:09
  • $\begingroup$ Would you mind posting this as an answer instead of a comment? If you do you can accept it and close the question as answered (so others don't answer the question when it's already been solved), and you can also collect upvotes and earn reputation. $\endgroup$ – Carl Schildkraut Jan 15 '19 at 5:23
  • $\begingroup$ Alright thanks for kindly advicing me! $\endgroup$ – Solvable Potato Jan 15 '19 at 5:32
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Hint: Recall Ceva's theorem:

If, in triangle $ABC$, points $D,E,F$ are on segments $BC,CA,AB$ respectively, then $AD,BE,CF$ concur if and only if

$$\frac{AF}{FB}\cdot\frac{BD}{DC}\cdot\frac{CE}{EA}=1.$$

Now, you know that, in triangle $ACY$, cevians $YB$, $AQ$, and $CP$ concur. What length ratios do you get when you apply Ceva's theorem? What length ratios would result in $PQ$ being parallel to $AC$?

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By menelaus's thm $$\frac{AC}{CB}\frac{BX}{XY}\frac{PY}{AP} = \frac{BC}{CA}\frac{BX}{XY}\frac{QY}{CQ} (=-1) \\hence\\ \frac{PY}{AP}=\frac{QY}{CQ} \\and\\ PQ//AC$$

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