Are there incongruent pythagorean triangles with the same perimeter and same area? I found there are two incongruent isosceles triangles with integer sides and areas, where both have same perimeter, same area.
I looked around Dickson's History of Number Theory but couldn't find where the right triangle version is treated. [I thought if a nonexistence proof was simple it would pop up in my search, but found none.] 
It may be simple to show none exist, but I had no luck, only filled  few notebook pages with formulas going nowhere. Reference/example/proof appreciated. Thanks.

Let $S$ be the circumference and $A$ twice the area of a triangle.
Then, 
$$a_i+b_i+\sqrt{a_i^2+b_i^2}=S
 \text{ and } a_ib_i=A. \tag{1}$$ 
After squaring, $S^2+2A-2S(a_i+b_i)=0$ and from here $a_i+b_i=S/2+A/S$. Thus, $c_i=S-(a_i+b_i)=S/2-A/S=c$, that is,  triangles have the same hypotenuse.
Then, $a_i+b_i=S-c=T$ and $a_ib_i=A$, which results in a solutions for $a_i$ and $b_i$ expressed in terms of constants $A$ and $T$. Although one of the resulting equations is quadratic, there is a symmetric pair of solutions (thank you for comments below). Hence, all sides must be the same.
