All triangles are isosceles

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 $P$.

Isosceles triangle

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.

Isosceles triangle 2

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 E. A. Maxwell, Fallacies in Mathematics, 1959. 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!

My very first lemma, which also happened to involve a philosophical dispute

Doppelverh-zentralproj.svgLet me recall the time of my very first lemma, which also happened to involve a philosophical dispute.

It was about 35 years ago; I was a kid in ninth grade at McKinley Junior High School, taking a class in geometry, taught by a charismatic math teacher. We were learning how to do proofs, which in that class always consisted of a numbered list of geometrical assertions, with a specific reason given for each assertion, either stating that it was “given” or that it followed from previous assertions by a theorem that we had come to know. Only certain types of reasons were allowed.

My instructor habitually used the overhead projector, writing on a kind of infinite scroll of transparency film, which he could wind up on one end of the projector, so as never to run out of room. During the semester, he had filled enough spools, it seemed, to fill the library of Alexandria.

One day, it came to be my turn to present to the rest of the class my proof of a certain geometrical theorem I had been assigned. I took the black marker and drew out my diagram and theorem statement. In my proof, I had found it convenient to first establish a certain critical fact, that two particular line segments in my diagram were congruent $\vec{PQ}\cong\vec{RS}$. In order to do so, I had added various construction lines to my diagram and reasoned with side-angle-side and so on.

Having established the congruency, I had then wanted to continue with my proof of the theorem. Since the previous construction lines were cluttering up my diagram, however, I simply erased them, leaving only my original diagram.

The class erupted with objection!  How could I sensibly continue now with my proof, claiming that $\vec{PQ}\cong\vec{RS}$, after I had erased the construction lines? After all, are those lines segments still congruent once we erase the construction lines that provided the reason we first knew this to be true? Many of the students believed that my having erased the construction lines invalidated my proof.

So there I was, in a ninth-grade math class, making a philosophical argument to my fellow students that the truth of the congruence $\vec{PQ}\cong\vec{RS}$ was independent of my having drawn the construction lines, and that we could rely on the truth of that fact later on in my proof, even if I were to erase those construction lines.

After coming to an uneasy, tentative resolution of this philosophical dispute, I was then allowed to continue with the rest of my proof, establishing the main theorem.

I realized only much later that this had been my very first lemma, since I had isolated a mathematically central fact about a certain situation, proving it with a separate argument, and then I had used that fact in the course of proving a more general theorem.