Relationship between Pythagorean Theorem and leg-hypotenuse lengths 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? 

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