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Given $ \Delta ABC$ is a triangle and $AD, BE$ and $CF$ are three concurrent lines. $P, Q$, and $R$ are points on $EF, FD, DE$ such that $ DP, EQ, FR$ are concurrent. Prove that $AP, BQ, CR$ are also concurrent.

I have invoked Ceva's theorem on the triangles but I have no idea what is to be done next.

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  • $\begingroup$ Aren't $D$, $E$, $F$ on the sides of $\triangle ABC$? Then it would be the Cevian nests theorem. $\endgroup$ – Sz_Z Oct 5 '17 at 18:44
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[EDIT: $BK$ should be parallel to $AP$, not $AC$. I've fixed it below.]

There may be an easier way, but here's how I managed to do it. I'll post just an outline of the idea for now and let you try to fill in the details:

Choose point $J$ on the line $AB$ so that $CJ \parallel EF$. Then choose point $K$ on the line $CJ$ so that $BK \parallel AP$. Finally let $AP$ intersect $BC$ and $CJ$ at points $P_1$ and $L$, respectively.

All these parallel lines produce several pairs of similar triangles (or alternatively, configurations where you can apply Thales' theorem). Try to express the ratio $\frac{CP_1}{P_1 B}$ in terms of other ratios in this way until you have shown it to be equal to $\frac{EP}{PF}\cdot\frac{AC}{AE}\cdot\frac{AF}{AB}$.

Then apply the same reasoning to the points where the other two lines, $BQ$ and $CR$, meet $CA$ and $AB$, respectively.

When you multiply them all, some things cancel, and two other groups are formed which become $1$'s by Ceva's theorem for the $\Delta ABC$ and $\Delta DEF$, respectively. Thus the whole product is $1$, which is what you needed to prove.

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