Assume that the two triangles $ABC$ and $A'B'C'$ have the common incircle and the common circumcircle. Let a point $P$ lie inside both the triangles. Prove that the sum of the distances from $P$ to the sidelines of triangle $ABC$ is equal to the sum of distances from $P$ to the sidelines of triangle $A'B'C'$.
Define the function $f$ mapping the Euclidean plane to the set of real numbers as follows: $f(X)$ is the sum of the oriented distances from $X$ to the sidelines of triangle $ABC$. Similarly, define the function $g$ for triangle $A'B'C'$. We need to prove that $f=g$.
Firstly, observe that $f$ is a linear function, i.e, if a point $R$ divides segment $PQ$ in the ratio $\lambda:1$ then $f(R)=\frac{f(P)+\lambda\cdot f(Q)}{1+\lambda}$. This can easily be seen by the Thales theorem.
Secondly, we consider the following Lemma.
Lemma The locus of all points $P$ such that $f(P)$ is a given constant is a line perpendicular to $OI$.
Proof See this
The same holds for $g$. Therefore, we only need to show that $f=g$ at $I$ and $O$. For $I$ the proof is obvious. We consider it for $O$.
Note that $f(O)=R\cdot (\cos A+\cos B+\cos C)=R\cdot (4\sin (A/2)\cdot \sin (B/2)\cdot sin(C/2)+1)=r+R=g(O)$. Hence the result holds.