Solving for triangle side length with limited information

Question

I have a geometry question which I am fairly certain has a solution, but I can't for the life of me figure out what it is.

I would like to find the length of $|\vec{AB}|$.

I have as givens:

1. $|\vec{AD}|$
2. $|\vec{BD}|$
3. $|\vec{AC}|$
4. $|\vec{AC}| = |\vec{BC}|$
5. $\vec{CD}$ is vertical
6. $\vec{AB}$ is vertical

I've added the line $\vec{EC}$ because I think it will be useful. I believe that there is a solution because when I draw this in a CAD program and enter the constraints it displays as fully defined, however I can't figure out how to calculate the length of AB myself.

Any suggestions or insights that you can provide are much appreciated. Thank you for reading!

Edited: Added that AB is vertical. I neglected to include that information in the original post.

Answer 1

Since the solution of a more general problem is equally easy, we drop the condition $AC=BC$. To simplify the notation we introduce: $$ AB=a,\quad AD=b,\quad BD=c,\quad AC=d,\quad BC=e. $$ Besides $$ \angle ADB=\theta_1,\quad \angle ACB=\theta_2. $$

For the trapezoid $ABCD$ we have: $$ \begin{align} &b^2+c^2-a^2=2bc\cos\theta_1;\\ &d^2+e^2-a^2=2de\cos\theta_2;\\ &bc\sin\theta_1=de\sin\theta_2. \end{align} $$ The first two equations is the cosine theorem, and the last one follows from the equality of the areas of the triangles $ACB$ and $ADB$.

Eliminating the angles one obtains the equation: $$ b^2c^2\left[1-\left(\frac{b^2+c^2-a^2}{2bc}\right)^2\right]= d^2e^2\left[1-\left(\frac{d^2+e^2-a^2}{2de}\right)^2\right], $$ which after straightforward algebra results in: $$ a^2=\frac{(b^2-c^2)^2-(d^2-e^2)^2}{2(b^2+c^2-d^2-e^2)}. $$

In your case the substitution $e=d$ will further simplify the result.