Square equals Sum of Squares implies Right Triangle

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Theorem

In the words of Euclid:

If in a triangle the square on one of the sides be equal to the squares on the other two sides of the triangle, the angle contained by the remaining two sides of the triangle is right.

(The Elements: Book $\text{I}$: Proposition $48$)


Proof

Euclid-I-48.png

Let $\triangle ABC$ be a triangle such that the square on $BC$ equals the sum of the squares on $AB$ and $AC$.

Construct $AD$ perpendicular to $AC$.

Make $AD$ equal to $AB$, and join $CD$.

Since $DA = AB$, the square on $AD$ equals the square on $AB$.

Add the square on $AC$ to each.

Then the squares on $AD$ and $AC$ equal the squares on $AB$ and $AC$.

But $\angle DAC$ is a right angle.

Therefore, by Pythagoras's Theorem, the square on $DC$ equals the sum of the squares on $AD$ and $AC$.

Also, the square on $BC$ equals the sum of the squares on $AB$ and $AC$, by hypothesis.

So the square on $DC$ equals the square on $BC$, and so $DC = BC$.

So from Triangle Side-Side-Side Congruence, $\triangle ABC = \triangle DAC$.

But as $\triangle DAC$ is a right triangle, then so must $\triangle ABC$ be.

$\blacksquare$


Historical Note

This proof is Proposition $48$ of Book $\text{I}$ of Euclid's The Elements.
It is the converse of Proposition $47$: Pythagoras's Theorem.


Sources

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