# Proof of converse of Pompeïu's Theorem

Pompeïu's Theorem states that for an equilateral $\Delta ABC$ and a point $P$ that is not on the circumcircle of the triangle, the lengths $PA, PB, PC$ form the sides of a triangle. Any such triangle is degenerate iff $P$ lies on the circumcircle.

The proof for this is simple enough, and can be obtained by a simple transformation. But, what about the converse? I was not able to find a proof for that anywhere.

EDIT: The converse states that if in a triangle $ABC$, for every point $P$ in its interior, $PB, PC, PA$ are the sides of a triangle, then $ABC$ is equilateral.

• Pompeiu's Theorem seems to be about points in the plane of a triangle, not just the interior. Be that as is may ... We can prove the contrapositive of the converse: "A non-equilateral triangle admits a point $P$ such that possibly-degenerate segments $\overline{PA}$, $\overline{PB}$, $\overline{PC}$ do not form a triangle." In the "plane" version, take $P$ to be a vertex where two non-congruent sides of the triangle meet. In the "interior" version, convince yourself that taking $P$ sufficiently close to that vertex creates segments that cannot form a triangle. – Blue Dec 15 '13 at 14:06
• @Blue: Yes, you are right. I should have mentioned that this is a slightly weaker version of the converse, so to speak. But it holds true, nonetheless. – Gerard Dec 15 '13 at 14:18
• RE Blue's comment. You can also take P close to the vertex but outside the triangle and not on the circumcircle ,so we get the full converse – DanielWainfleet Nov 1 '15 at 15:57 We can prove for any none equilateral $ΔABC$ we will find a point $K$ that KA,KB,KC can't form a triangle.
WOLG, let $BC$ is max side,$AB$ is min side, find $CD=AB, E$ is midpoint of $DB$, take $F$ on $EB$,make a red circle with $r=FB$, and get $GB=FB$ on $AB$. take H on $GB$. line $CH$, corss red circle at $I$, take $K$ on $IH$,
then $AK<AB=CD,KB<FB=ED \implies AK+KB<CD+EB=CE$,
but $CK>CI>CF>CE >AK+KB$