If we allow for the degenerate case, then the triangle of side lengths 0 satisfies all the conditions mentioned in the problem, but I'm assuming that's not the solution you are looking for / I need to prove only the degenerate case works.
Let x,y,z be the lengths of the right angle triangle with z as the hypotenuse
Let a,a,b be the lengths of the isosceles triangle.
We have that $\displaystyle x+y+z=2a+b$ if we assume perimeters must be equal
$\displaystyle x^2+y^2=z^2$ by Pythagoras
$\displaystyle p(p-x)(p-y)(p-z) = p(p-a)(p-a)(p-b)$ by Heron's formula (squaring both sides) since areas are equal. We can use p here on both sides since perimeters are equal and hence the semi-perimeter is equal as well.
If p=0, then we have the degenerate case, so assume p is not equal to 0 and divide both sides by p
so,
$\displaystyle (p-x)(p-y)(p-z) = (p-a)(p-a)(p-b)$
$\displaystyle p^3 -(x+y+z)p^2 + (xy + xz + yz)p - xyz = p^3 - (2a+b)p^2 + (a^2 + 2ab)p - a^2b$
Hence by comparison we have
$\displaystyle x+y+z = 2a+b$ (i)
$\displaystyle xy+xz+yz = a^2+2ab$ (ii)
$\displaystyle xyz = a^2b$ (iii)
By multiplying (ii) by x and using (iii), we get
$\displaystyle x^2y + x^2z + xyz = a^2x + 2abx$
$\displaystyle x^2 [y+z] = a^2(x-b) + 2abx$ (1)
Similarily
$\displaystyle y^2 [x+z] = a^2(y-b) + 2aby$ (2)
$\displaystyle z^2 [x+y] = a^2(z-b) + 2abz$ (3)
I have to use Pythagoras eventually, but it feels like I'm going in circles.
Maybe I'm going about this question incorrectly. Initially I wanted to use the concept for height as an argument. In the xyz right triangle, if I choose x as the base y is the height. I'm thinking about what happens if I draw an isosceles triangle with the base on the same level as the base of the xyz right triangle. Areas are equal if base lengths are equal and height in the same. hmm...