Proving inequalities between areas 
In triangle $ABC$ the bisector from $A$ intersects the circumcircle at $A_1$, and let $B_1$ and $C_1$ be defined the same way. Let $AA_1$ intersect the outer angle bisectors of angles $ABC$ and $ACB$ at $A_0$, and let $B_0$ and $C_0$ be defined the same way. Prove that $[A_0 B_0 C_0]= 2[A C_1 B A_1 C B_1]\geq 4[ABC]$ where $[k]$ means the area of $k$.

I think I have solved it but I am not sure. I got to the point of showing that all of the triangles are equilateral and then got the result easy. I wonder, if we assume that $x$ $>=$ $y$ $>=$ $z$ $<=>$ $2x$ $>=$ $2y$ $>=$ $2z$, where $x$, $y$ and $z$ are angles of a triangle and $x+y$, $y+z$, $z+x$ are angles of another triangle then we get: $x+y$ $>=$ $2y$, $y+z$ $>=$ $2z$ and $z+x$ $<=$ $2x$ then we get $z$ $>=$ $x$ which contradicts our assumption so must therefore all angles be equal? Thank you responses
 A: This is an interesting exercise in "role exchanging".



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*The points $A_1,B_1,C_1$ are the midpoints of the $BC,AC,AB$ arcs in the circumcircle of $ABC$

*The points $A_0,B_0,C_0$ are the excenters of $ABC$

*If we denote through $H$ the incenter of $ABC$ we have that $H$ is the orthocenter of $A_0 B_0 C_0$ and $ABC$ is the orthic triangle of $A_0 B_0 C_0$

*In particular the circumcircle of $ABC$ is the nine-point-circle of $A_0 B_0 C_0$ and $A_1,B_1,C_1$ are the midpoints of $HA_0,HB_0,HC_0$ $\ldots$

*And $A_1 B_1 C_1$ and $A_0 B_0 C_0$ are both homothetic and perspective, with their areas having a ratio of $\frac{1}{4}$

*Since $A_1 B_1$ is the perpendicular bisector of $CH$ and so on, the area of the hexagon $AC_1 B A_1 C B_1$ is exactly twice the area of $A_1 B_1 C_1$ (it is enough to split the hexagon as the union of three kites with a common vertex at $H$).

*By Euler's theorem for pedal triangles the area of the pedal triangle of $P$ in $A_0 B_0 C_0$ just depends on the distance of $P$ from the circumcenter of $A_0 B_0 C_0$. In particular the maximum area of a pedal triangle is the area of the pedal triangle of the circumcenter, i.e. the area of the medial triangle. This proves $[ABC]\leq \frac{1}{4}[A_0 B_0 C_0]$ and finishes the proof.


We have $[ABC]=\frac{1}{4}[A_0 B_0 C_0]$ if and only if the orthocenter and the circumcenter of $A_0 B_0 C_0$ are the same point, i.e. if and only if $A_0 B_0 C_0$ is equilateral, i.e. if and only if $ABC$ is equilateral.
