Costruct the externaly equilateral triangle on BC with vertex A'', on CA with vertex B'' and on AB with vertex C''. with an homothety with center A we have that A, A' and A'' are align and equal for B,B'B'' and C,C',C''. Then AA', BB' CC' councur on the First Fermat Point.

Virgil Nicula wrote: An easy extension. Let $ ABC$ be an acute triangle. Consider the isosceles triangle $ A'UV$ , where $ A' \in (BC)$ , $ U\in (AC)$ , $ V\in (AB)$ such that $ m(\widehat {A'UV}) = m(\widehat {A'VU}) = \phi$ (constant) and $ UV \parallel BC$ . Define the points $ B',C'$ in the same way. Prove that $ AA'$, $ BB'$ and $ CC'$ are concurrently . Proof. Prove easily that $ \frac {A'B}{A'C} = \frac {1 + \tan\phi \cdot\cot B}{1 + \tan\phi\cdot\cot C}$ . Indeed, denote the midpoint $ M$ of $ [UV]$ , $ MU = MV = x$ and the projections $ N$, $ P$ of $ V$ , $ U$ on $ BC$ respectively. Observe that $ A'N = A'P = x$ , $ MA' = VN = UP = x\tan\phi$ and $ BN = x\tan\phi\cot B$ , $ PC = x\tan\phi\cot C$ . Thus, $ \left\|\begin{array}{c} A'B = A'N + NB = x + x\tan\phi\cot B \\ \ A'C = A'P + PC = x + x\tan\phi\cot C\end{array}\right\|$ $ \implies$ $ \frac {A'B}{A'C} = \frac {1 + \tan\phi\cot B}{1 + \tan\phi\cot C}$ . Obtain analogously $ \frac {B'C}{B'A} = \frac {1 + \tan\phi \cdot\cot C}{1 + \tan\phi\cdot\cot A}$ and $ \frac {C'A}{C'B} = \frac {1 + \tan\phi \cdot\cot A}{1 + \tan\phi\cdot\cot B}$ . Now apply the Ceva's theorem. Remark. The midpoint $ M$ of the segment $ UV$ belongs to the $ A$ - median $ [AD]$ , where $ D\in (BC)$ and his position is given by the ratio $ \frac {MA}{MD} = \frac {2h_a}{a\tan\phi}$ . Prove easily that $ \frac {1}{UV} = \frac 1a + \frac {\tan\phi}{2h_a}$ .

Other solution Apply Ceva theory for triagle $ AUV$ : $ \frac {\sin{\angle {BAA'}}}{\sin{\angle {CAA'}}}.\frac {\sin{\angle {A'UV}}}{\sin{\angle {A'UA}}}.\frac {\sin{\angle {A'VA}}}{\sin{\angle{UVA'}}} = 1$ Note that $ \angle {A'UV} = \angle {A'VU} = 60^{o}$ a $ \angle {AUA'} = B + 60^{o}$ $ \angle {AVA'} = C + 60^{o}$ So $ \prod \frac {\sin{\angle {BAA'}}}{\sin{\angle {CAA'}}} = 1$ So $ AA',BB',CC'$ are concurrent .

I think that it 's not so true. Since we have 2 ways of constructing A' and similarly 2 ways of constructing B', C'. If we restrict ourself to constructing only either external or internal equilateral triangles then it is true. If we mix them (construct BCA'' externally and ABC'', CAB'' internally) then it is false.

Virgil Nicula wrote: An easy extension. Let $ ABC$ be an acute triangle. Consider the isosceles triangle $ A'UV$ , where $ A' \in (BC)$ , $ U\in (AC)$ , $ V\in (AB)$ such that $ m(\widehat {A'UV}) = m(\widehat {A'VU}) = \phi$ (constant) and $ UV \parallel BC$ . Define the points $ B',C'$ in the same way. Prove that $ AA'$, $ BB'$ and $ CC'$ are concurrently . Too much gun power to kill a bee . Consider the triangles $ A'UV$ as in hypothesis and $ B'XY$, $ C'ZT$ used to define the points $ B'$ and $ C'$, respectively, where $ B' \in (AC)$ , $ X\in (BC)$ , $ Y\in (AB)$, $ C' \in (AB)$ , $ Z\in (AC)$ , $ T\in (BC)$. Then, the following triangles are similar (provided that $ \phi$ is the same in all three constructions and using the parralelism of $ UV$ and $ BC$, $ XY$ and $ AC$, $ ZT$ and $ AB$, respectively): $ VA'C$ and $ XB'C$, implying $ \frac {A'C}{CB'} = \frac {A'V}{B'X}$ $ YB'A$ and $ ZC'A$, implying $ \frac {B'A}{AC'} = \frac {B'Y}{C'Z}$ $ TC'B$ and $ UA'B$, implying $ \frac {C'B}{BA'} = \frac {C'T}{A'U}$ It results that: $ \frac {A'C}{CB'}\times\frac {B'A}{AC'}\times\frac {C'B}{BA'} = \frac {A'V}{B'X}\times\frac {B'Y}{C'Z}\times\frac {C'T}{A'U} = 1$, provided that the three triangles $ A'UV$, $ B'XY$ and $ C'ZT$ are isosceles. From Ceva Th., it results that $ AA'$, $ BB'$ and $ CC'$ are concurrent.