Question: Every point in a plane is assigned some real number. It is found that for any triangle, the number at its incenter is the arithmetic mean of numbers at the vertices. Prove that every point has been assigned the same number.
Solution: Let us take any two random points $A$ and $D$ on the real plane $\mathbb{R}^2$. Now let us select two points $B$ and $C$ such that $AD||BC, AD<BC$ and $AB=DC$. Also let the real numbers assigned to $A,B,C,D$ be $a,b,c,d$ respectively.
Observe that $ABCD$ is an isosceles trapezium with $AB=DC$.
Now let us produce $BA$ and $CD$ to meet at $E$.
Observe that in $\Delta EAD$, we have $\angle EAD=\angle EDA\implies EA=ED$. Thus we also have $EB=EC$. Hence, $\Delta EAD$ and $\Delta EBC$ are both isosceles triangles with $EA=ED$ and $EB=EC$ respectively.
Let the perpendicular dropped from $E$ to $BC$ meet $BC$ at $F$.
Now consider $\Delta EBD$. $EF$ is the angle bisector of $\angle BED$ and let that the angle bisector of $\angle EBD$ meet $EF$ at $I$. Thus $I$ is the incentre of $\Delta EBD$.
Now consider $\Delta ECA$. $EF$ is the angle bisector of $\angle AEC$. Join $C$ and $I$. Observe that $\Delta EBI$ and $\Delta ECI$ are congruent triangles, which implies that $\angle EBI=\angle ECI=x$ (let).
This implies that $\angle EBI=\angle IBD=x$, which in turn implies that $\angle EBD= 2x$.
Again, $\Delta ADB \cong \Delta DAC\implies \angle DBA=\angle ACD\implies 2x=x+\angle ICA\implies \angle ICA=x.$ Hence, $\angle ICD=\angle ICA=x$, which implies that $CI$ is the angle bisector of $\angle ECA$.
Hence, we can conclude that $I$ is also the incenter of $\Delta ECA$.
Hence the number assigned to $$I=\frac{b+d+e}{3}=\frac{a+c+e}{3}.$$ Thus we have $b+d=c+a$.
Thus, we can conclude that for any random isosceles trapezium $ABCD$ with $AD||BC$ and $AD<BC$ having real numbers $a,b,c,d$ assigned to vertices $A,B,C,D$ respectively, we will have $$a+c=b+d.$$ Let this property be called $\phi$.
Now select any two random points $A'$ and $B'$ on the real plane $\mathbb{R}^2$ and construct a regular pentagon $A'B'C'D'E'$. Let the real number assigned to $A',B',C',D',E'$ be $a',b',c',d',e'$ respectively.
Now $B'C'E'D'$ is an isosceles trapezium, hence by $\phi$, we have $b'+d'=e'+c'$. Similarly, $e'+c'=a'+d'$, $a'+c'=b'+d'$ and $b'+e'=a'+d'$. From these four equations we can conclude that $a'=b'=c'=d'=e'$.
Thus we have $a'=b'$. From here, we can conclude that, since $A$ and $B$ were randomly chosen, therefore, we can reach to the same conclusion, i.e, $a'=b'$, with any other pair of points $\in\mathbb{R}^2$.
Hence, we can conclude that every point on the real plane $\mathbb{R}^2$ has been assigned the same number.
Is this solution correct and is there a better solution than this one?