Find four unknown catheti lengths from three configured hypotenuses Given this configuration of right-angled triangles as pictured, with only the provided information available — that is, the complete dimensions of the lesser triangle plus the two hypotenuse lengths of the other triangles.

I need to be able to calculate formulaically the unknown dimensions for the other two triangles. I have a whole series of these configurations that have decimalised catheti and are not in neat proportions hence why I need to find the formulae.
Please help as I have spent quite a bit of time trying to work this out without any success.
As for my efforts I have little to show for my time spent on this problem.
I can see that there are orthogonal pairs of parallel lines comprising the catheti.
I can also see the simplistic formulae like, for example AF = AE + 200 and AE = AF - 200, but none of these formulae are calculable without further information (I think) or bring me to the values.
For my mind, if I could just find the formula for calculating the length of BE then everything else is solvable by way of the dashed rectangle.
What is the length of BE?
 A: I have managed to find the solution:
∠CBA = arccos(((AB² + BC² - AC²)/(2•AB•BC)) = 158.5221008351°
∠CBD = 26.8701104101°
∠DBE = 90°
∠EBA = ∠CBA - ∠CBD - ∠DBE = 41.6519904251°
BE = AB•cos(∠EBA) = 185.5923415927
To find the remaining side lengths use variations of @dxiv helpful answers to find the rest:
BE = DF
AE = √(AB² - BE²)
AF = 200 + AE
CF = CD + BE
A: Expanding on my comments, this can be solved without trigonometry by using Pythagora's theorem for right triangles $\triangle ABE$ and $\triangle ACF\,$:
$$
\begin{align}
\begin{cases}
AE^2+BE^2 &= AB^2
\\ (AE+EF)^2+(BE+CD)^2 &= AC^2
\end{cases}
\end{align}
$$
Expanding the latter and using the former:
$$
\begin{align}
AC^2 &= AE^2+ 2\cdot AE\cdot EF + EF^2+BE^2+2\cdot BE\cdot CD + CD^2
\\ &= \underbrace{AE^2+BE^2}_{\displaystyle=\;AB^2}+EF^2 + CD^2+2\cdot BE\cdot CD+ 2\cdot AE\cdot EF 
\\ \implies\;\;\;\; 2\cdot AE \cdot EF&= AC^2-AB^2-CD^2-EF^2-2\cdot BE\cdot CD
\end{align}
$$
Substituting back into the first equation $AE^2+BE^2 = AB^2$ gives a quadratic in $BE\,$:
$$
\left(AC^2-AB^2-CD^2-EF^2-2\cdot BE\cdot CD\right)^2 + 4 \cdot BE^2 \cdot EF^2 = 4 \cdot AB^2 \cdot EF^2
$$
With $AB \approx 248.4, AC \approx 464.3, CD \approx 101.3, EF = 200$ the equation is (approximately):
$$
4 \cdot 200^2 \cdot BE^2 + \left(464.3^2-248.4^2-101.3^2-200^2 - 2 \cdot 101.3 \cdot BE\right)^2-4\cdot248.4^2\cdot 200^2=0
$$
The solutions are $\,BE \approx 185.7\,$ and $\,BE \approx 23.1\,$ (which can be calculated with better accuracy by using the full precision of the given lengths).
The fact that there are two solutions indicates that the construction is not unique with just the given constraints. Indeed, the triangle $\triangle BCD$ is univocally determined by the lengths of its sides, but point $A$ is defined by its distances to $B$ and $C$, and there are two matching points $A$ which are the two intersections of the circles centered at $B$ with radius $BA \approx 248.4\,$, and respectively centered at $C$ with radius $CA\approx 464.3\,$.
A: Thank you once again, @dxiv. This is even more helpful to me. I just couldn’t make that link until you spelt it out. Much appreciated. It so happens that last night I found this same geometrical double point relationship.

