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

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.
