If you want to convince someone that certain triples are the side lengths of a right triangle, you can exhibit pictures like the following, which show the $(4,3,5)$, $(12,5,13)$, and $(8,15,17)$ right triangles.
The idea here is to let the given triple be $(a,b,c)$. Then one of the diagonals of the tilted rectangle is vertical and has length $2c$, while half of the other diagonal, which is on a slant, coincides with the hypotenuse of the right triangle with legs $a$ and $b$. Since the diagonals of a rectangle are equal, this demonstrates that the hypotenuse of the right triangle has length $c$.
Some details: the corners of the tilted rectangle are at $(0,c)$, $(a,b)$, $(0,-c)$, and $(-a,-b)$. You know this is a rectangle because the slope of the line joining the first and second points is $-\frac{c-b}{a}$, while the slope of the line joining the second and third points is $\frac{c+b}{a}$. These are negative reciprocals of each other because the equality
$$
-\left(-\frac{c-b}{a}\right)^{-1}=\frac{a}{c-b}=\frac{c+b}{a}
$$
is equivalent to the Pythagorean theorem. Although we are pretending we don't know the Pythagorean theorem, the given triple must satisfy it nonetheless. At this point, the negative reciprocal property is just be a property that happens to be satisfied by the given numbers. The corners of the right triangle are at $(0,0)$, $(0,b)$, and $(a,b)$.
This is actually only a few steps shy of a proof of the Pythagorean theorem, as long as one is willing to countenance use of properties of similar triangles and a bit of algebra.
Given $\triangle ACB$ with opposite sides $a$, $c$, and $b$, extend the line through $\overline{AC}$ and draw the circle with center $A$ and radius $\overline{AB}$, intersecting the line at $D$ and $E$. The rectangle is now easily constructed. Since $\triangle DCB$ and $\triangle BCE$ are similar, we have the result,
$$
\frac{c-b}{a}=\frac{a}{c+b}.
$$
If one wishes to stick with purely geometric reasoning, construct three squares. Construct $ACJK$ on $\overline{AC}$ (area is $b^2$). Construct $ADLM$ on $\overline{AD}$ (area is $c^2$). Then, drawing the perpendicular to $\overline{CB}$ through $B$ and the perpendicular to $\overline{DE}$ through $E$, the two perpendiculars meeting at $F$, subdivide rectangle $BCEF$ by a line $\overline{GH}$ into a square $EFGH$ (area is $a^2$) and a rectangle $BCHG$. The Pythagorean theorem is the statement that the gnomon (L-shaped region) $CDLMKJ$ and the square $EFGH$ have equal area.
This can be proved by cut-and-paste geometry with the aid of the line perpendicular to $\overline{GH}$ through the point $I$ where $\overline{GH}$ and $\overline{BE}$ intersect. Let this line intersect $\overline{CB}$ at $N$ and $\overline{EF}$ at $P$. Then rectangle $CNPE$ equals the gnomon in area. Since this rectangle's area equals the area of $\triangle BCE$ plus the area of $\triangle EPI$ minus the area of $\triangle BNI$, while the area of square $EFGH$ equals the area of $\triangle EFB$ plus the area of $\triangle IHE$ minus the area of $\triangle IGB$, and corresponding triangles in these calculations are congruent in pairs, the result is established.
