Easy by complex bush

Let $ABC$ be the initial triangle $T$, and let $XYZ$, with $X \in BC, Y \in AC, Z \in AB$ be the triangle emerging from the intersection of lines of the same colour. Finally, let $O$ be the circumcenter of triangle $ABC$. Note that $\angle OXC=30^\circ=\angle OYA,$ and so quadrilateral $OYCX$ is cyclic, hence $\dfrac{XY}{\sin \angle XCY}=\dfrac{OC}{\sin \angle OXC}=2OC=\dfrac{AB}{\sin \angle C},$ and since $\angle XCY \equiv \angle C,$ we obtain that $XY=AB$. In a similar fashion $YZ=BC$ και $ZX=AC$, hence we are evidently done.

Label the points in the problem as shown in the diagram. Then as $$\measuredangle AOA'=\measuredangle AP_bA'=\measuredangle AP_cA'=120^{\circ}$$we have that $AA'OP_bP_c$ is cyclic and it has equal radius to the circumcircle of $T$. Thus as $P_bP_c$ and $BC$ inscribe an equal arc length we have $P_bP_c=BC$ and cyclic variants suffice to prove the problem.

Attachments:

The final triangle is given by rotating the medial triangle around the circumcenter and then dilating by a factor of $2$, which finishes.