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.

Consider two right triangles, with hypotenuses $p$ and $q$ and respective acute angles $\theta$ and $\phi$. To see that having equal perimeter and area makes them congruent, it suffices to show that $$\theta = \phi \qquad\text{or}\qquad \theta+\phi=\frac{\pi}{2} \tag{0}$$ (Either makes the triangles similar, which in turn makes them congruent.)

Equating perimeters and areas gives a system we can write as

$$\begin{align} p(1+\sin\theta+\cos\theta) &= q(1+\sin\phi+\cos\phi) \\ p^2 \sin\theta \cos\theta &= q^2 \sin\phi \cos\phi \end{align} \tag{1}$$

Defining $u:=\tan(\theta/2)$, we "know" that $$\sin\theta = \frac{2u}{1+u^2} \qquad \cos\theta=\frac{1-u^2}{1+u^2} \quad\to\quad 1+\cos\theta+\sin\theta= \frac{2 (1 + u)}{1 + u^2}$$ and likewise for $v:=\tan(\phi/2)$. Thus, $(1)$ can be rewritten as $$\begin{align} p\frac{(1+u)}{1+u^2} &= q\frac{(1+v)}{1+v^2} \\[4pt] p^2\frac{u(1+u)(1-u)}{(1+u^2)^2} &= q^2\frac{v(1+v)(1-v)}{(1+v^2)^2} \end{align}\tag{2}$$ Dividing the second equation by the square of the first ...

$$\frac{u(1-u)}{1+u} = \frac{v(1-v)}{1+v} \quad\to\quad (u-v)(uv+u+v-1)=0 \quad\to\quad u=v, \text{ or } \frac{u+v}{1-uv}=1 \tag{3}$$

Therefore, we have one of the following situations (bearing in mind that $\theta/2$ and $\phi/2$ are each at most $\pi/4$, so that we may draw appropriate conclusions from these tangent inequalities): $$\begin{align} \tan\frac{\theta}{2}=\tan\frac{\phi}{2} &\quad\to\quad \theta=\phi \\[4pt] \frac{\tan(\theta/2)+\tan(\phi/2)}{1-\tan(\theta/2)\tan(\phi/2)} = 1 &\quad\to\quad \tan\left(\frac{\theta}{2}+\frac{\phi}{2}\right)=\tan\frac{\pi}{4} \quad\to\quad \theta+\phi=\frac{\pi}{2}\end{align} \tag{4}$$ which match the sufficient conditions in $(0)$. $\square$

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.