Let me share a mathematical gem with you, the following paradoxical “theorem.”

**Theorem.** Every triangle is isosceles.

**Proof.** Consider an arbitrary triangle $\triangle ABC$. Let $Q$ be the intersection of the angle bisector (blue) at $\angle A$ and the perpendicular bisector (green) of $BC$ at midpoint $P$.

Drop perpendiculars from $Q$ to $AB$ at $R$ and to $AC$ at $S$. Because $P$ is the midpoint of $BC$ and $PQ$ is perpendicular, we deduce $BQ\cong CQ$ by the Pythagorean theorem. Since $AQ$ is the angle bisector of $\angle A$, the triangles $AQR$ and $AQS$ are similar, and since they share a hypotenuse, they are congruent. It follows that $AR\cong AS$ and also $QR\cong QS$. Therefore $\triangle BQR$ is congruent to $\triangle CQS$ by the hypotenuse-leg congruence theorem. So $RB\cong SC$. And therefore,

$$AB\cong AR+RB\cong AS+SC\cong AC,$$

and so the triangle is isosceles, as desired. **QED**

**Corollary.** Every triangle is equilateral.

**Proof.** The previous argument proceeded from an arbitrary vertex of the triangle, and so any pair of adjacent sides in the triangle are congruent. So all three are congruent, and therefore it is equilateral. **QED**

Perhaps you object to my figure, because depending on the triangle, perhaps the angle bisector of $A$ passes on the other side of the midpoint $P$ of $BC$, which would make the point $Q$ lie outside the triangle, as in the following figure.

Nevertheless, essentially the same argument works also in this case. Namely, we again let $Q$ be the intersection of the angle bisector at $\angle A$ with the perpendicular bisector of $BC$ at midpoint $P$, and again drop the perpendiculars from $Q$ to $R$ and $S$. Again, we get $BQ\cong CQ$ by the Pythagorean theorem, using the green triangles. And again, we get $\triangle ARQ\cong\triangle ASQ$ since these are similar triangles with the same hypotenuse. So again, we conclude $\triangle BQR\cong\triangle CQS$ by hypotenuse-leg. So we deduce $AB\cong AR-BR\cong AS-CS\cong AC$, by subtracting rather than adding as before, and so the triangle is isosceles.

**Question.** What is wrong with these arguments?

Please post your answers in the comments below.

The argument is evidently due to W. W. Rouse Ball, *Mathematical Recreations and Essays* (1892). I first heard it years ago, when I was in graduate school. Shortly afterward, my advisor W. Hugh Woodin happened to be a little late to seminar, and so I leaped to the chalkboard and gave this proof, leaving the distinguished audience, including R. Solovay, scratching their heads for a while. Woodin arrived, but Solovay wouldn’t let him start the seminar, since he wanted to resolve the triangle argument. What fun!