The longer the base, the longer the hypotenuse Excuse me if this is a silly question, but my plane geometry is very rusty. When I re-read Jack D'Aurizio's answer to the question "How can we prove that $\pi > 3$ using this definition", I wondered why, when viewed from the perspective of high school plane geometry, the area of the regular hexagon of unit spoke length is smaller than the area of the unit disc. Obviously, this follows if the regular hexagon entirely lies inside the unit circle, but why is this true? My first thought was that this is because the distance from the centre of the hexagon to any point on its perimeter is at most $1$. Put it another way, this is a consequence of the following (stronger) statement:


*

*Given $\bigtriangleup ABC$, if $X$ lies on the line segment $\overline{BC}$, then the length of $\overline{AX}$ is $\le$ the larger length of $\overline{AB}$ or $\overline{AC}$.



The above statement in turn follows from another statement:


*

*Let $\overline{AP}$ be a fixed line segment and $\bigtriangleup APX$ be a right-angled triangle, where $\overline{AP}\perp\overline{PX}$. The longer the base $\overline{PX}$, the longer the hypotenuse $\overline{AX}$.


This follows directly from Pythagoras theorem. However, all proofs of Pythagoras theorem that I knew, such as the ancient Chinese proof or Einstein's proof, make use of the notion of area. Why must the proof of a statement about length involve the concept of area?
So, here is my question:


*

*Can the statement in the second bullet point be proved, within the framework of Eucliedan plane geometry, without using Pythagoras theorem or the notion of area?

 A: One of Euclid's useful propositions is that the lengths of sides are in the same order as the opposite angles. I.e., in $\triangle PQR$ we have $PQ>PR$ if and only if $\angle R>\angle Q$. (This follows from the fact that an exterior angle of a triangle is greater than either remote interior angle.) Most students these days understand this in terms of the law of sines, but it's far more elementary.
So, in your picture, take $X'$ between $B$ and $X$, so that $PX'>PX$. Consider $\triangle AXX'$. Then you can conclude $AX'>AX$ because $\angle X>\angle X'$.
A: Migrating a comment to an answer, as requested.

Proving the Pythagorean Theorem doesn't require calculating area. Suppose, for the purpose of this comment, that your $\triangle ABC$ has a right angle at $A$. Then $\overline{AP}$ creates similar triangles, and we have
$$\frac{|\overline{PB}|}{|\overline{AB}|}=\frac{|\overline{AB}|}{|\overline{BC}|} \qquad \qquad
\frac{|\overline{PC}|}{|\overline{AC}|}=\frac{|\overline{AC}|}{|\overline{BC}|}$$
so that
$$|\overline{AB}|^2 + |\overline{AC}|^2 =
|\overline{BC}||\overline{PB}|+|\overline{BC}||\overline{PC}|=
|\overline{BC}|\;\left(\;|\overline{PB}|+|\overline{PC}|\;\right) = |\overline{BC}|^2$$ 
A: Let extend $PX$ and take point $X_1$ such as $PX_1>PX$. Then it's easy to see that $\angle{AXX_1}$ is obtuse and $AX_1$ is the biggest side in $\triangle{AXX_1}$
A: for Ted's answer, these are pages 168 and 198 from Marvin Jay Greenberg, Euclidean and Non-Euclidean Geometry, 4th edition. In the third edition it was pages 121 and 142. 


A: Note:
$$\tan \measuredangle PXA=\frac{AP}{PX}; \\
\sin \measuredangle PXA = \frac{AP}{AX}; \\
0<\measuredangle PXA<90.$$
When $\measuredangle PXA$ decreases, both $\sin \measuredangle PXA$ and $\tan \measuredangle PXA$ will decrease, hence both $PX$ and $AX$ will increase (since $AP$ is fixed).
