Is there any polygon exists such that sum of length of square of two adjacent sides is equal to another side/diagonal?

Question

In Right angle triangle we have $ a^2 + b^2 = c^2$
where $a^2 = (x_1-x_2)^2 + (y_1-y_2)^2 ,$
$b^2 = (x_3-x_2)^2 + (y_3-y_2)^2 $
and $c^2 = (x_1-x_3)^2 + (y_1-y_3)^2$

And in Square we have

$ a^2 + b^2 = c^2$
$ d^2 + e^2 = c^2$
$ a^2 + d^2 = f^2$
$ b^2 + e^2 = f^2$
and $a=b=d=e , c=f$
where $a^2 = (x_1-x_2)^2 + (y_1-y_2)^2 ,$
$b^2 = (x_3-x_2)^2 + (y_3-y_2)^2 $ ,
$c^2 = (x_1-x_3)^2 + (y_1-y_3)^2$ ,
$d^2= (x_1-x_4)^2+(y_1-y_4)^2$ ,
$e^2= (x_3-x_4)^2+(y_3-y_4)^2$ ,
and $f^2= (x_2-x_4)^2+(y_2-y_4)^2$

Is there any other polygon exists which posses this property (sum of length of square of two adjacent sides is equal to another side/diagonal ) ?
It's a sufficient condition to say if this property is satisfied by a polygon then it's a right angle triangle or square ?

Answer 1

Two adjacant sides and the corresponding diagonal fulfill this equality iff they form a right triangle. Some simple examples of such polygons are polyominoes, or any polygon with only horizontal and vertical edges. If you allow the "$c$" to be an unrelated diagonal or side, many many other polygons are possible.

Answer 2

Perhaps you're looking for something like the hexagon $(3,4,12,84,3612,3613)$ where $3^2+4^2+12^2+84^2+3612^2=3613^2$. Polygons like this are found by matching side $C$ of one triple with side $A$ of another. There are infinite combinations with the number of sides ranging from $3$-to-$\infty$. I can show you formulas for finding them if you like.

There are also tiles composed of other triples with alternately matching sides B-with-B and C-with-C such as the isosceles triangle below. Little can be said of it except that the sums squares of sides adjacent to right angles add up to the squares of the sides opposite the right angles.