Denote $BC = a$, $CA = b$, $AB = c$. The assumptions imply $c \le a$, $c \le b$. First we prove that $b + c = 2a$, without using the condition that $ABC$ is a right triangle.

Let $F$ be the midpoint of $BE$. By $BD \perp DE$ triangle $BED$ is right at $D$, so $DF$ is the median to its hypotenuse $BE$. Hence
$$BF = DF = EF = \frac{1}{2} BE.$$
On the other hand $AB = AD$, so $AF$ and $F$ are equidistant from the endpoints of $BD$. Hence $AF$ is the perpendicular bisector of $BD$. Because triangle $BDA$ is isosceles with base $BD$, it follows that $AF$ is the bisector of $A$.

By the bisector property $\frac{AB}{BF} = \frac{AC}{CF}$. Replacing $AB = c$, $BF = \frac{c}{2}$, $AC = b$, $CF = a - \frac{c}{2}$ yields
$$\frac{c}{2} = \frac{b}{a - \frac{c}{2}}.$$
In particular $BC = a$ is the middle side of $ABC$, and since $AB = c$ is the shortest one, the hypotenuse of the triangle is $AC = b$. Thus $b^2 = a^2 + c^2$ by Pythagoras theorem. Combined with $b = 2a - c$ this yields $(2a - c)^2 = a^2 + c^2$, which reduces to $3a = 4c$. Hence $a = 4d$, $c = 3d$ with
$$d > 0, \text{ and } b = \sqrt{a^2 + c^2} = 5d.$$