The longer the base, the longer the hypotenuse

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 $\angle X>\angle X'$.

Migrating a comment to an answer, as requested.

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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 $$|\overline{AB}|^2 + |\overline{AC}|^2 = |\overline{BC}||\overline{PB}|+|\overline{BC}||\overline{PC}|= |\overline{BC}|\;\left(\;|\overline{PB}|+|\overline{PC}|\;\right) = |\overline{BC}|^2$$

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 $\triangle{AXX_1}$