This is a well known problem (see the links below). It's also interesting to note that there are more remarkable points P, besides the Fermat points, whose cevians AP,BP,CP reflected about BC,CA,AB concur, namely the points P lying on the Isocubic K060 with pivot X(265) and pole X(1989). http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=304719 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=175130

[geogebra]144ec5f21f51a926bb17a3b94d6ef9aaacdc61a1[/geogebra] They concurrent at one of the isodynamic points , which is one the intersections of 3 Apollonius circles(each circle has takes the intersection of internal/external bisectors and the respective side of $\triangle ABC$ as a diameter , you can refer to here), construct the isodynamic point first and then prove each symmetric line passes through it, by and ratio method you like, I did with lots of trigonometry calculations which I don't think is proper to post ^_^ You may see basic features(with proofs) of the isodynamic points in this pdf file

Notice that $ AT $ bisects $ \angle BTC $ , consider the isogonal conjugate of $A$ w.r.t. $ \Delta BTC $ , $P$ , we have$ \angle BCP = \angle ACT = \angle BCT^* ~,~ \angle CBP = \angle ABT = \angle CBT^* $ , where $ T^*$ is the isogonal conjugate of $T$ w.r.t. $ \Delta ABC$ , implying that $ P$ is the reflection of $T^*$ in $BC$ but we find that $ A , T , P $ are collinear , the reflection of $ AT$ in $ BC$ is therefore a line passing through $ T^*$ .

Notice that it is equivalent to saying that reflections of $T$-the Fermat point in $BC,CA,AB$ when joined with $A,B,C$ respectively concur. This is equivalent to $T$ being on the Neuberg cubic which is well-known.