Let D', E' and F' be points on BC, CA and AB respectively such that BD = CD' ; BD' = CD , CE = AE' ; CE' = AE and AF = BF' ; AF' = BF. (Reflections over midpoints of the sides of the triangle ABC) These give BD/DC = CD'/D'B ; CE/EA = AE'/E'C ; AF/FB = BF'/F'A. Further, with information from the problem, CD'/D'B = CE/EA (BD/DC = AE'/E'C) ; BF'/F'A = BD/DC (AF/FB = CD'/D'B) ; AE'/E'C = AF/FB (CE/EA = BF'/F'A). These give DF'//CA, D'F//CA ; ED'//AB, E'D//AB ; FE'//BC, F'E//BC. Let ∠CAB = A ; ∠ABC = B ; ∠BCA = C. Let O be the circumcenter of ∆ABC and ∆DEF. By the definition of D' the midpoint of DD' is same point as the midpoint of BC, hence the perpendicular bisector of BC is the same as the perpendicular bisector of DD' since the points are all collinear. Therefore, the perpendicular bisector of DD' passes through O as well as the perpendicular bisector of DE since it is the circumcenter of ∆DEF, hence, O is also the circumcenter of ∆DD'E. Similarly, O is also the circumcenter of ∆EE'F and ∆FF'D. Therefore, the perpendicular bisectors of DD', D'E, EE', E'F, FF' and F'D all pass through O. This implies D, D', E, E', F and F' are concyclic points. Now, DF'//CA and F'E//BC, we have CDF'E is a parallelogram, therefore, ∠DF'E = ∠BCA = C, and ∠ED'C = ∠ABC = B from ED'//AB. Also DD'EF' is a cyclic quadrilateral, therefore, ∠DF'E = ∠ED'C, hence, B = C. Also, DF'//CA and E'D//AB, we have AF'DE' is a parallelogram, therefore, ∠F'DE' = ∠CAB = A and ∠E'FA = ∠CAB = B from FE'//BC. Also FF'DE' is a cyclic quadrilateral, therefore, ∠F'DE' = ∠E'FA, hence, A = B. Therefore, A = B = C, hence the triangle ABC is equilateral.

Usually analytic solutions to a problem turn to be quite lengthy and messy, but in this case they turn out not to be too bad. With the circumcentre of $\triangle ABC$ as the origin, we have that there is a real number $t$ such that $$ \begin{array}{ccccc} \vec{D} = \vec{B} + t(\vec{C} - \vec{B}) & \text{and} & \vec{E} = \vec{C} + t(\vec{A} - \vec{C}) & \text{and} & F = \vec{A} + t(\vec{B} - \vec{A}) \end{array} $$ Since the origin is the circumcentre of $\triangle DEF$, we have that $|\vec{D}|^2 = |\vec{E}|^2 = |\vec{F}|^2$. But $$ |\vec{D}|^2 = (1 - t)^2 |\vec{B}|^2 + 2t(1 - t) \vec{B} \cdot \vec{C} + t^2 |\vec{C}|^2 $$and we can find a similar formula for $|\vec{E}|^2$ and $|\vec{F}|^2$. Equating these and remembering that $|\vec{A}|^2 = |\vec{B}|^2 = |\vec{C}|^2 = R^2$ where $R$ is the circumradius of $\triangle ABC$, we find that $$ \vec{B} \cdot \vec{C} = \vec{C} \cdot \vec{A} = \vec{A} \cdot \vec{B} $$which implies that $$ \cos\angle BOC = \cos\angle COA = \cos\angle AOB $$which implies that $\triangle ABC$ is equilateral. A proof using coordinates or complex numbers would be similar.

Yet another solution is given by using Stewart's Theorem. Let $O$ be the common circumcentre of $\triangle ABC$ and $\triangle DEF$, and let $BD = a$ and $DC = ka$. Stewart's Theorem in $\triangle BOC$ give us that $$ (k + 1)a (OD^2 + ka^2) = R^2 ka + R^2 a $$where $R = OB = OC$ is the circumradius of $\triangle ABC$. We thus have that $ka^2 = R^2 - OD^2$, and a similar calculation gives us that if $CE = b$ and $AF = c$, then $ka^2 = kb^2 = kc^2 = R^2 - r^2$ where $r = OD = OE = OF$ is the circumradius of $\triangle DEF$. It follows that $a = b = c$, and hence that $AB = BC = CA$.

Can we say that for the conditions stated, the centroid of $\triangle ABC$ and $\triangle DEF$ will always coincide? Then if we say that if this is so, then having the circumcentres of $\triangle ABC$ and $\triangle DEF$ coincide means that the circumcentre must coincide with its centroid, hence both triangles are equilateral?

Power of point of $D$ to the circuncircle of ABC is $DB\times DC$ and $-DO^2 +R^2$. The power of point of $E$ is $EC\times EA$ and $ -EO^2 +R^2$. As $EO=DO$, the product of the lengths is the same. Together with the stated condition of ratios, gives us that $DC=EA$ and $DB=EC$. Analogously for the others pairs of sides the triangle is equilateral

Let the radius of the circumcircle of $ABC$ be $r$ and let the circumcenter of $ABC$ be $O$. Clearly, $AO=OB=r$. Now, let $BD = a, DC = b, EC = ac, AE = bc, AF=ad,$ and $FB = bd$. By Stewart’s theorem on $BOC$, we have that: $$(a)(b)(a+b) + OD^2(a+b) = r^2(a+b) \implies ab + OD^2 = r^2 \implies OD^2 = r^2-ab$$By the same reasoning, $$OE^2 = r^2-bac^2$$and $$OF^2 = r^2-abd^2$$Since $OE=OF=OD$, $abd^2 = abc^2=ab \implies d = c = 1$, and $ABC$ is equilateral.

Mmesomachi wrote: Let $D'$, $E'$ and $F'$ be points on $BC$, $CA$ and $AB$ respectively such that $BD = $; $BD' = CD$ , $CE = AE' ; CE' = AE$ and $AF = BF' ; AF' = BF$. (Reflections over midpoints of the sides of the triangle $ABC$) These give $\frac{BD}{DC}= \frac{CD'}{D'B} ; \frac{CE}{EA}= \frac{AE'}{E'C}; \frac{AF}{FB} = \frac{BF'}{F'A}$. Further, with information from the problem, $\frac{CD'}{D'B}= \frac{CE}{EA }$ $$\frac{BD}{DC} =\frac{AE'}{E'C} ; \frac{BF'}{F'A} = \frac{BD}{DC} ;\frac{AF}{FB} =\frac{CD'}{D'B} ; \frac{AE'}{E'C} = \frac{AF}{FB} ; \frac{CE}{EA}= \frac{BF'}{F'A}$$These give $DF'//CA, D'F//CA ; D'//AB, E'D//AB ; FE'//BC, F'E//BC$. Let $\angle CAB = A ; \angle ABC = B ; \angle BCA = C$. Let $O$ be the circumcenter of $ABC$ and $DEF$. By the definition of $D'$ the midpoint of $DD'$ is the same point as the midpoint of $BC$, hence the perpendicular bisector of $BC$ is the same as the perpendicular bisector of $DD'$ since the points are all collinear. Therefore, the perpendicular bisector of $DD'$ passes through $O$ as well as the perpendicular bisector of $DE$ since it is the circumcenter of $DEF$, hence, $O$ is also the circumcenter of $DD'E$. Similarly, $O$ is also the circumcenter of $EE'F$ and $FF'D$. Therefore, the perpendicular bisectors of $DD'$, $D'E, EE', E'F, FF'$ and $F'D$ all pass through $O$. This implies $D, D', E, E', F$ and $F'$ are concyclic points. Now, $DF'//CA$ and $F'E//BC$, we have $CDF'E$ is a parallelogram, therefore, $\angle DF'E = \angle BCA = C$, and $\angle ED'C = \angle ABC = B$ from $ED'//AB$. Also $DD'EF'$ is a cyclic quadrilateral, therefore, $\angle DF'E = \angle ED'C$, hence, $B = C$. Also, $DF'//CA$ and $E'D//AB$, we have $AF'DE'$ is a parallelogram, therefore, $\angle F'DE' = \angle CAB = A$ and $\angle E'FA = \angle CAB = B$ from $FE'//BC$. Also $FF'DE'$ is a cyclic quadrilateral, therefore, $\angle F'DE' = \angle E'FA$, hence, $A = B$. Therefore, $A = B = C$, hence the triangle $ABC$ is equilateral.

DylanN wrote: Yet another solution is given by using Stewart's Theorem. Let $O$ be the common circumcentre of $\triangle ABC$ and $\triangle DEF$, and let $BD = a$ and $DC = ka$. Stewart's Theorem in $\triangle BOC$ give us that $$ (k + 1)a (OD^2 + ka^2) = R^2 ka + R^2 a $$where $R = OB = OC$ is the circumradius of $\triangle ABC$. We thus have that $ka^2 = R^2 - OD^2$, and a similar calculation gives us that if $CE = b$ and $AF = c$, then $ka^2 = kb^2 = kc^2 = R^2 - r^2$ where $r = OD = OE = OF$ is the circumradius of $\triangle DEF$. It follows that $a = b = c$, and hence that $AB = BC = CA$. This is cool