I have the following figure:
Let's consider two right triangles, with right angles at points $L$ and $M$, respectively. Since the lines passing through points $A$ and $C$ and through points $L$ and $M$ are parallel, intersected by a transversal passing through points $A$ and $L$, alternate interior angles are equal. Therefore, triangles ALC and AML are similar, hence we have $\frac{LM}{AM}=\frac{AL}{LC}$. Consequently, $\frac{LM}{AM}$ is known to be $k\in\mathbb R$ and the side $LC$ is known to be $a\in\mathbb R$, while the unknown is the hypotenuse $AC$. A straightforward calculation yields $AC=a\sqrt{k^2+1}$.
Initially, making an error, I had erroneously deduced from the first equality that $AL=k$ instead of $AL=k\cdot a$, which sparked a small curiosity. I wonder, how should the above diagram be modified to have $AL=k$, and thus $AC=\sqrt{k^2+a^2}$? I'm struggling to visualize the modified figure, but perhaps this question lacks interest or is poorly framed? What are your thoughts?