I have known the following for 45 years: in the Euclidian plane, every triangle is isosceles.
The false proof needs a handmade picture; take your pen, it's easy. Start from a triangle $ABC$. Draw the perpendicular bisector of $BC$, and the angle bisector from $A$. Let $I$ be their intersection (if it is not unique, you are done). Let $J$ be the projection of $I$ over $AB$, $K$ that over $AC$. Considering the right triangles $AIJ$ and $AIK$, we see that (lengths) $AJ=AK$, and that $IJ=IK$. Then looking at righttriangles $BIJ$ and $CIK$, we obtain that $BJ=CK$. We conclude that $$AB=AJ+JB=AK+KC=AC.$$
The falsity is that one of $J$ or $K$ is in the triangle, and the other one is out. Therefore one of the sums above (and only one) should be a difference.