Relationship between Pythagorean Theorem and leg-hypotenuse lengths

Question

With a standard right triangle, we have the legs $a$ and $b$ and the hypotenuse $c$, where typically $a$ is the shorter leg. I'm curious about the fact that

$$a + b > c$$

but

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

This discussion seems to be a simple proof -- assuming the Pythagorean Theorem is true. However, at an intuitive level, it seems odd that what wasn't equal is made equal by squaring each member. Is there any theoretical explanation from higher geometry?

Answer 1

In $\triangle ABC$ with $\angle C$ denoting the interior angle at the vertex $ C$ and the sides' lengths being $a,b,c$ we have the Cosine Law: $$c^2=a^2+b^2-2ab\cdot \cos \angle C$$ which is very ancient. Therefore $$(a+b)^2=a^2+b^2+2ab=$$ $$=a^2+b^2-2ab \cdot \cos \angle C+2ab(1+\cos \angle C)=$$ $$=c^2+2ab (1+\cos \angle C).$$ Since $0<\angle C<\pi$ we have $1+\cos \angle C>0.$ Therefore $$(a+b)^2=c^2+2ab(1+\cos \angle C)>c^2$$ which implies $a+b>c.$

The fact that the interior angles of a triangle sum to $\pi\;$ ( implying $\angle C<\pi\;$ ) follows from the Parallel Postulate. The Cosine Law follows directly from the Theorem of Pythagoras, which follows from the Parallel Postulate, and the Theorem of Pythagoras is proved without needing to know that $a+b>c.$

The Theorem of Pythagoras can be considered a special case of the Cosine Law when $\angle C=\pi /2$ and $\cos \angle C=0.$ We can re-arrange the Cosine Law as $$\cos \angle C=\frac {a^2+b^2-c^2}{2ab}$$ from which we can see that if $\angle C<\pi /2$ then $a^2+b^2>c^2,$ and if $ \angle C>\pi /2$ then $a^2+b^2<c^2.$

Answer 2

In Geometry there is an important difference between the triangle inequality $a+b>c$ and Pythagoras' theorem $a^2+b^2=c^2$: the proof of the former doesn't need the Parallel Postulate, while the proof of the latter does.

As a matter of fact, in Euclid's Elements the triangle inequality is Prop. 20 of Book I, while Pythagoras' theorem is proved much later, as Prop. 47 of Book I.