Theorem: 3 Proper Integer Heronian Triangles whose Area and Perimeter are Equal

There are exactly $3$ proper integer Heronian triangles whose area and perimeter are equal.
These are the triangles whose sides are:
:$\tuple {6, 25, 29}$
:$\tuple {7, 15, 20}$
:$\tuple {9, 10, 17}$

Proof:
First, using Pythagoras's Theorem, we establish that these integer Heronian triangles are indeed proper:
{{begin-eqn}}
{{eqn | l = 6^2 + 25^2
      | r = 661
      | c = 
}}
{{eqn | o = \ne
      | r = 29^2
      | c = so not right-angled
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 7^2 + 15^2
      | r = 274
      | c = 
}}
{{eqn | o = \ne
      | r = 20^2
      | c = so not right-angled
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 9^2 + 10^2
      | r = 181
      | c = 
}}
{{eqn | o = \ne
      | r = 17^2
      | c = so not right-angled
}}
{{end-eqn}}
Now we show they have area equal to perimeter.
We use Heron's Formula throughout:
:$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
where:
:$\AA$ denotes the area of the triangle
:$a$, $b$ and $c$ denote the lengths of the sides of the triangle
:$s = \dfrac {a + b + c} 2$ denotes the semiperimeter of the triangle.
Thus we take the $3$ triangles in turn:
{{begin-eqn}}
{{eqn | n = 6, 25, 29
      | l = s
      | r = \frac {6 + 25 + 29} 2
      | c = 
}}
{{eqn | r = 30
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }
      | r = \sqrt {30 \paren {30 - 6} \paren {30 - 25} \paren {30 - 29} }
      | c = 
}}
{{eqn | r = \sqrt {30 \times 24 \times 5 \times 1}
      | c = 
}}
{{eqn | r = \sqrt {3600}
      | c = 
}}
{{eqn | r = 60
      | c = 
}}
{{eqn | r = 6 + 25 + 29
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 7, 15, 20
      | l = s
      | r = \frac {7 + 15 + 20} 2
      | c = 
}}
{{eqn | r = 21
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }
      | r = \sqrt {21 \paren {21 - 7} \paren {21 - 15} \paren {21 - 20} }
      | c = 
}}
{{eqn | r = \sqrt {21 \times 14 \times 6 \times 1}
      | c = 
}}
{{eqn | r = \sqrt {1764}
      | c = 
}}
{{eqn | r = 42
      | c = 
}}
{{eqn | r = 7 + 15 + 20
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 9, 10, 17
      | l = s
      | r = \frac {9 + 10 + 17} 2
      | c = 
}}
{{eqn | r = 18
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }
      | r = \sqrt {18 \paren {18 - 9} \paren {18 - 10} \paren {18 - 17} }
      | c = 
}}
{{eqn | r = \sqrt {18 \times 9 \times 8 \times 1}
      | c = 
}}
{{eqn | r = \sqrt {1296}
      | c = 
}}
{{eqn | r = 36
      | c = 
}}
{{eqn | r = 9 + 10 + 17
      | c = 
}}
{{end-eqn}}
It remains to be demonstrated that these are indeed the only such proper integer Heronian triangles which match the criterion.
Let $\tuple {a, b, c}$ be the sides of such a triangle.
Using Heron's Formula, we have:
{{begin-eqn}}
{{eqn | l = 2 s
      | r = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }
}}
{{eqn | ll= \leadsto 
      | l = 4 s^2
      | r = s \paren {s - a} \paren {s - b} \paren {s - c}
}}
{{eqn | ll= \leadsto
      | l = 4 s
      | r = \paren {s - a} \paren {s - b} \paren {s - c}
}}
{{end-eqn}}
Note that:
:$\paren {s - a} + \paren {s - b} + \paren {s - c} = 3 s - a - b - c = s$
Hence by substituting $x = s - a$, $y = s - b$, $z = s - c$:
:$4 \paren {x + y + z} = x y z$
By Semiperimeter of Integer Heronian Triangle is Composite, $s$ is an integer.
Hence $s, x, y, z \in \N_{>0}$.
By Triple with Product Quadruple the Sum, our equation has solutions:
:$\tuple {1, 5, 24}, \tuple {1, 6, 14}, \tuple {1, 8, 9}, \tuple {2, 3, 10}, \tuple {2, 4, 6}$
Using:
:$a = s - x = x + y + z - x = y + z$
:$b = s - y = x + z$
:$c = s - z = x + y$
the possible sets of side lengths are:
:$\tuple {29, 25, 6}, \tuple {20, 15, 7}, \tuple {17, 10, 9}, \tuple {13, 12, 5}, \tuple {10, 8, 6}$
of which the final $2$ are Pythagorean Triples, so they are not proper Heronian triangles.
Hence the result.
{{qed}}