parmenides51 wrote: $QF\parallel BC$ I believe this should be $QF\parallel AC$ Let $H$, $K$ be points on $AB$, $AC$ such that $FH \perp AB$, $FK \perp AC$, $M$ be the intersection of $FP$ and $AC$, $N$ be the intersection of $FQ$ and $AB$. Obviously $AMFN$ is a parallelogram. We have $\angle NQA = 90^{\circ} - \angle BPQ = 90^{\circ} - \angle APB = \angle LAB = \angle NAQ$ $\Rightarrow NQ = NA = MF$. Similarly, $NF = MP$, so $\triangle NHF = \triangle MB_1P$ and $\triangle NC_1Q = \triangle MKF$. It is easy to see that $\triangle AC_1Q \sim \triangle AB_1P$, so $\frac{FH}{FK} = \frac{PB_1}{QC_1} = \frac{AB_1}{AC_1} = \frac{AB}{AC}$. Therefore $F$ must lies on the $A-symmedian$ of $\triangle ABC$, so $\overline{A, F, T}$

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