Length of a pivoted rectangle

Question

I have a rectangle with a swivel point along the left edge at the midpoint between the top and bottom of the rectangle. The swivel point will be the point "a", the bottom right corner is point "b", the top right corner is point "d" and the point on the right edge equidistant between point "b" and "d" is point "f". I need to know the length from point "a" to point "f" under the follow constraints as the rectangle is pivoted upwards with point "a" being fixed. The original length of the rectangle is unknown but there are some known endpoints.

As the rectangle pivots upwards a triangle will be formed with points "a" "b" and "c". Point "c" is below the rectangle on the line that line segment "af" was originally on prior to pivoting up. The vertex "acb" is a 90-degree angle and the length between points "a" and "c" is known. Along line segment "ac" is point "e". Point "e" forms a triangle with points "a" and "d".The vertex "aed" is a 90-degree angle and the length of "ed" is known. How do I find the length between points "a" and "f"?

I've tried to look at all the triangles formed from the lines. I know if I figured out a single triangle everything else would fall into place. With the parallel lines that the points fall on the problem feels like there should be some known relationships but I can't figure it out.

Answer 1

Let $h$ be the intersection between $af$ and $de$, and set: $A=ac$, $B=de$, $C=df$, $X=af$, $\theta=\angle fae$. We have then: $$ ae=ac-ec=A-2C\sin\theta;\quad eh=ed-dh=B-{C\over\cos\theta};\quad ah=af-fh=X-C\tan\theta. $$ But $eh=ah\sin\theta$ and $ae=ah\cos\theta$, hence we get two equations for the unknowns $X$ and $\theta$: $$ B=X\sin\theta+C\cos\theta;\quad A=C\sin\theta+X\cos\theta. $$ The problem is then reduced to solve the above coupled equations for $X$, but the task doesn't seem so trivial: I tried with Mathematica obtaining a very large expression which cannot be simplified.

EDIT.

Eliminating $\theta$ from the above equations one gets a single quartic equation in X: $$ X^4-(A^2 + B^2 + 2C^2)X^2 + 4ABCX + C^2(C^2- A^2 - B^2) =0. $$ But a general formula for the solution is very large. Better to substitute there the values of $A$, $B$, $C$ and solve numerically.