TITLE: Proving a theorem using its converse
QUESTION [21 upvotes]: Converse of Pythagoras' theorem: If the lengths of the sides of a triangle $T$ are $a$, $b$, and $c$, and if $a^2+b^2=c^2$, then the triangle is a right triangle and the side opposite to the right angle is the one whose length is $c$.
Proof: Construct a line segment $XY$ whose length is $a$. Then construct a line segment $YZ$ whose length is $b$ which is perpendicular to $XY$. By construction, the triangle $XYZ$ is a right triangle, and therefore, by Pythagoras' theorem and because we are assuming that $a^2+b^2=c^2$, the length of $XZ$ is equal to $c$. So, the triangle $XYZ$ is similar to the original triangle $T$. Since the triangle $XYZ$ is a right triangle, then so is $T$.What I find peculiar about this proof is the fact that it uses Pythagoras' theorem in order to prove its converse.
It is not the only situation that I am aware of in which this occurs. For instance, there is a proof of the converse of Ceva's theorem  which uses that theorem. But I am not aware of any example outside Euclidean Geometry.

Can anyone provide an example of a theorem of the type $A\implies B$ outside Geometry with a proof which uses the fact that $B\implies A$?

REPLY [2 votes]: Definition: Graph $G$ is perfect if each $H\leq G$ has chromatic number the same as clique number.
Theorem: (László Lovász, 1972) Graph $G$ is perfect iff it complement $\overline{G}$ is perfect.
Proof:

Suppose if part is proven.
Now let $\overline{G}$ be perfect. Then by already proven part also $\overline{\overline{G}}$ is perfect. But $\overline{\overline{G}} =G$ and we are done.