Interesting property of triangle and ellipses formed on its edges I need to prove following theorem. It seems to work, and it seems to be intuitive but I think I am missing skills in geometry proofs to prove that.
Clearer version of theorem: For any triangle $ABC$ on a plane, we draw circle $O$ such that $AB$ is its diameter. Placing a point $P$ inside $O$ will result in creating triangle (either $PAB$, $PBC$ or $PAC$) of smaller perimeter than $ABC$.
Previous (base theorem) version: For triangle on a plane, we can draw an ellipse based on each of its edges such that end points of that edge are focal points and common point of two other edges lies on edge of an that ellipse (it describes every triangle with same "base" edge and every combination of two other edges such that they form triangle with same perimeter as first one). We can draw 3 such ellipses for triangle - let's call their union $E$. Prove that circles circumscribed on each edge (edge is diameter of circle) of that triangle lie inside of $E$.
Where did it come from?
I've found an interesting answer on problem of finding 3 points from set of points such that triangle built of them has as small perimeter as possible  - it's here. I want to prove that this method is working, as it's not "clear" for me at all. As it's easier to prove anything on Delaunay triangulation by referencing Voronoi diagrams I tried that method - with no good results.
Theorem above is a bit more than this, but it seems to work (used visualization software to check it and couldn't create counterexample).
EDIT: Here are examples. As you can see, without losing a generality, we choose $AB$ edge and draw circle such that $AB$ is its diameter. Now this circle is always inside union of ellipses.


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*Example 2

*Example 3
 A: Take a point R on the circle with diameter AB. Then the opposite point $C_0$ yields three ellipses such that R is on the boundary. Note that $A$, $B$, $C_0$, $R$ form a rectangle. Now if we call $d(A,B)=AB$, then 
$$
RA+RB=BC_0+AC_0,
$$
hence $R$ is on the same ellipsis with focal points $A$ and $B$ as $C_0$.
Moreover $R$ is in the interior of each ellipsis with focal points $A$ and $B$, that passes through $C$, if 
$$
RA+RB<BC+AC.
$$
This means that all $C$ outside the ellipsis
$$
RA+RB=BX+AX,
$$
is in the good zone $Z_1$. 
Moreover 
$$
RA+RC_0=BA+BC_0,
$$
hence $R$ is on the same ellipsis with focal points $A$ and $C_0$ which passes trough $B$.
Moreover $R$ is in the interior of each ellipsis with focal points $A$ and $C$, that passes through $B$, if 
$$
RA+RC<BA+BC.
$$
This means that all $C$ on one side of a branch
of an hiperbola given by
$$
BX-RX=RA-AB
$$
(which passes through $C_0$)
is in the good zone $Z_2$.
Now note that 
$$
RB+RC_0=AB+AC_0,
$$
hence $R$ is on the same ellipsis with focal points $B$ and $C_0$ which passes trough $A$.
Moreover $R$ is in the interior of each ellipsis with focal points $A$ and $C$, that passes through $B$, if 
$$
RB+RC<BA+AC.
$$
This means that all $C$ on one side of a branch
of an hiperbola given by
$$
AX-RX=RB-AB
$$
(which passes through $C_0$)
is in the good zone $Z_3$.
Finally one can prove that the union of $Z_1$, $Z_2$ and $Z_3$ is the whole cartesian plane,
hence for each $R$ and all $C$, one of the ellipsis defined by $A$, $B$ and $C$ contain $R$. Since $R$ is arbitrary, this proves the theorem.
