# Proof of the Converse of the Pythagorean Theorem

The Pythagorean Theorem states that if $ABC$ is a triangle right angled at $C$, then,

$a^2 + b^2 = c^2$.

The converse of the Pythagorean Theorem states that if

$a^2 + b^2 = c^2$ holds,

then triangle $ABC$ is a right triangle right angled at $C$.

This means that in order to prove the converse of the Pythagorean Theorem, we need to prove that $\angle C$ in the figure above is a right angle.  Now, we discuss the proof.

Theorem

In a triangle with sides $a$, $b$ and $c$ (see figure above), if

$a^2 + b^2 = c^2$ holds,

then $ABC$ is a right triangle with a right angle at $C$.

Proof

Let $DEF$ be a triangle such that $EF = a$, $DF = b$ and right angled at $F$ (see figure below). If we let $DE = x$, since $DEF$ is a right triangle, by the Pythagorean Theorem

$a^2 + b^2 = x^2$ (1).

But from the supposition,

$a^2 + b^2 = c^2$ (2).

From (1) and (2)

$x^2 = c^2$.

Since $x$ and $c$ are both positive (Can you see why?), we can therefore conclude that

$x = c$.

This means that length of the three corresponding pairs of sides of triangle $ABC$ and triangle $DEF$ are equal.

Therefore, by SSS Congruence, $\triangle ABC \cong \triangle DEF$.

Since $F$ and $C$ are corresponding angles, $\angle F = \angle C = 90$ degrees.

And hence we have proved that triangle $ABC$ is right angled at $C$

* Note that in the proof, $a$ is the opposite of angle $A$, $b$ is the opposite of angle $B$, and $c$ is the opposite side of angle $C$.