# Proof of the Converse of the Pythagorean Theorem

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

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

The converse of the Pythagorean Theorem states that if

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

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

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

Theorem

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

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

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

Proof

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

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

But from the supposition,

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

From (1) and (2)

\$latex x^2 = c^2\$.

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

\$latex x = c\$.

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

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

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

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

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