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 $latex C$.

right triangle

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.


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 $latex C$.


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).

right triangle 3.

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 $latex C$.

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